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STUDY PACKAGE
Subject : Mathematics
Topic: Permutation &Combination
Index
1. Theory
2. Short Revision
3. Exercise (1 to 5)
4. Assertion & Reason
5. Que. from Compt. Exams
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Permutation and Combination
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Permutations are arrangements and combinations are selections. In this chapter we
discuss the methods of counting of arrangements and selections. The basic results
and formulas are as follows:
(i)
Principle of Multiplication:
If an event can occur in ‘m’ different ways, following which another event can occur in ‘n’
different ways, then total number of different ways of simultaneous occurrence of both
the events in a definite order is m × n.
(ii)
Principle of Addition:
If an event can occur in ‘m’ different ways, and another event can occur in ‘n’ different
ways, then exactly one of the events can happen in m + n ways.
Example # 1
There are 8 buses running from Kota to Jaipur and 10 buses running from Jaipur to
Delhi. In how many ways a person can travel from Kota to Delhi via Jaipur by bus.
Solution.
Let E1 be the event of travelling from Kota to Jaipur & E2 be the event of travelling from
Jaipur to Delhi by the person.
E1 can happen in 8 ways and E2 can happen in 10 ways.
Since both the events E1 and E2 are to be happened in order, simultaneously, the number
of ways = 8 × 10 = 80.
Example # 2
How many numbers between 10 and 10,000 can be formed by using the digits 1, 2, 3,
4, 5 if
(i)
No digit is repeated in any number.
(ii)
Digits can be repeated.
Solution.
(i)
Number of two digit numbers = 5 × 4 = 20
Number of three digit numbers = 5 × 4 × 3 = 60
Number of four digit numbers = 5 × 4 × 3 × 2 = 120
Total = 200
(ii)
Number of two digit numbers = 5 × 5 = 25
Number of three digit numbers = 5 × 5 × 5 = 125
Number of four digit numbers = 5 × 5 × 5 × 5 = 625
Total = 775
Self Practice Problems :
1.
How many 4 digit numbers are there, without repetition of digits, if each number is
divisible by 5.
Ans. 952
2.
Using 6 different flags, how many different signals can be made by using atleast three
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Fundamental Principle of Counting :
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1.
Pr = n (n − 1) (n − 2)..... (n − r + 1) =
NOTE : (i) factorials of negative integers are not defined.
(ii) 0 ! = 1 ! = 1 ;
(iii) nPn = n ! = n. (n − 1) !
(iv) (2n) ! = 2n. n ! [1. 3. 5. 7... (2n − 1)]
Example # 3
How many numbers of three digits can be formed using the digits 1, 2, 3, 4, 5, without
repetition of digits. How many of these are even.
Solution.
Three places are to be filled with 5 different objects.
∴
Number of ways = 5P3 = 5 × 4 × 3 = 60
For the 2nd part, unit digit can be filled in two ways & the remaining two digits can be
filled in 4P2 ways.
∴
Number of even numbers = 2 × 4P2 = 24.
Example # 4
If all the letters of the word 'QUEST' are arranged in all possible ways and put in dictionary
order, then find the rank of the given word.
Solution.
Number of words beginning with E = 4P4 = 24
Number of wards beginning with QE = 3P3 = 6
Number of words beginning with QS = 6
Number of words beginning withQT = 6.
Next word is 'QUEST'
∴
its rank is 24 + 6 + 6 + 6 + 1 = 43.
Self Practice Problems :
3.
Find the sum of all four digit numbers (without repetition of digits) formed using the
digits 1, 2, 3, 4, 5.
Ans. 399960
4.
Find 'n', if n – 1P3 : nP4 = 1 : 9.
Ans. 9
5.
Six horses take part in a race. In how many ways can these horses come in the first,
second and third place, if a particular horse is among the three winners (Assume No
Ties).
Ans. 60
3.
Circular Permutation :
The number of circular permutations of n different things taken all at a time is; (n − 1) !.
If clockwise & anti−clockwise circular permutations are considered to be same, then it
is
(n − 1)!
.
2
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n!
(n − r )!
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n
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2.
flags, arranging one above the other.
Ans. 1920
Arrangement :
If nPr denotes the number of permutations of n different things, taking r at a time, then
Number of circular permutations of n things when p alike and the rest different taken all at a time
distinguishing clockwise and anticlockwise arrangement is
(n −1)!
.
p!
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Note:
Example # 6
In how many ways 6 persons can sit at a round table, if two of them prefer to sit together.
Solution.
Let P1, P 2, P3, P4, P5, P 6 be the persons, where P1, P2 want to sit together.
Regard these person as 5 objects. They can be arranged in a circle in (5 – 1)! = 24. Now P1P2 can be
arranged in 2! ways. Thus the total number of ways = 24 × 2 = 48.
98930 58881
Example # 5
In how many ways can we arrange 6 different flowers in a circle. In how many ways we can form a
garland using these flowers.
Solution.
The number of circular arrangements of 6 different flowers = (6 – 1)! = 120
When we form a garland, clockwise and anticlockwise arrangements are similar. Therefore, the number
1
(6 – 1) ! = 60.
of ways of forming garland =
2
6.
In how many ways the letters of the word 'MONDAY' can be written around a circle if the vowels are to
be separated in any arrangement.
Ans. 72
7.
In how many ways we can form a garland using 3 different red flowers, 5 different yellow flowers and 4
different blue flowers, if flowers of same colour must be together.
Ans. 17280.
4.
Selection :
If nCr denotes the number of combinations of n different things taken r at a time, then
n
Cr =
n
n!
P
= r where r ≤ n ; n ∈ N and r ∈ W.
r! (n − r )!
r!
NOTE : (i) nCr = nCn – r
(ii) nCr + nCr – 1 = n + 1Cr
(iii) nCr = 0 if r ∉ {0, 1, 2, 3........, n}
Example # 7
Fifteen players are selected for a cricket match.
(i)
In how many ways the playing 11 can be selected
(ii)
In how many ways the playing 11 can be selected including a particular player.
(iii)
In how many ways the playing 11 can be selected excluding two particular players.
Solution.
(i)
11 players are to be selected from 15
Number of ways = 15C11 = 1365.
(ii)
Since one player is already included, we have to select 10 from the remaining 14
Number of ways = 14C10 = 1001.
(iii)
Since two players are to be excluded, we have to select 11 from the remaining 13.
Number of ways = 13C11 = 78.
Example # 8
If 49C3r – 2 = 49C2r + 1, find 'r'.
Solution.
n
Cr = nCs if either r = s or r + s = n.
Thus 3r – 2 = 2r + 1
⇒
r=3
or
3r – 2 + 2r + 1 = 49
⇒
5r – 1 = 49
⇒
r = 10
∴
r = 3, 10
Example # 9
A regular polygon has 20 sides. How many triangles can be drawn by using the vertices, but not using
the sides.
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Self Practice Problems :
20 × (17 C 2 − 16 )
= 800.
3
Example # 10
10 persons are sitting in a row. In how many ways we can select three of them if adjacent persons are
not selected.
Solution.
Let P1, P2, P3, P4, P5, P6, P 7, P8, P 9, P10 be the persons sitting in this order.
If three are selected (non consecutive) then 7 are left out.
Let PPPPPPP be the left out & q, q, q be the selected. The number of ways in which these 3 q's can
be placed into the 8 positions between the P's (including extremes) is the number ways of required
selection.
Thus number of ways = 8C3 = 56.
Example # 11
In how many ways we can select 4 letters from the letters of the word MΙSSΙSSΙPPΙ.
Solution.
M
ΙΙΙΙ
SSSS
PP
Number of ways of selecting 4 alike letters = 2C1 = 2.
Number of ways of selecting 3 alike and 1 different letters = 2C1 × 3C1 = 6
Number of ways of selecting 2 alike and 2 alike letters = 3C 2 = 3
Number of ways of selecting 2 alike & 2 different = 3C1 × 3C2 = 9
Number of ways of selecting 4 different = 4C4 = 1
Total = 21
Self Practice Problems :
8.
In how many ways 7 persons can be selected from among 5 Indian, 4 British & 2 Chinese, if atleast two
are to be selected from each country.
Ans. 100
9.
10 points lie in a plane, of which 4 points are collinear. Barring these 4 points no three of the 10 points
are collinear. How many quadrilaterals can be drawn.
Ans. 185.
10.
In how many ways 5 boys & 5 girls can sit at a round table so that girls & boys sit alternate.
Ans. 2880
11.
In how many ways 4 persons can occupy 10 chairs in a row, if no two sit on adjacent chairs.
Ans. 840.
12.
In how many ways we can select 3 letters of the word PROPORTION.
Ans. 36
5.
The number of permutations of 'n' things, taken all at a time, when 'p' of them are similar & of one type,
q of them are similar & of another type, 'r' of them are similar & of a third type & the remaining
n − (p + q + r) are all different is
n!
.
p! q! r !
Example # 12
In how many ways we can arrange 3 red flowers, 4 yellow flowers and 5 white flowers in a row. In how
many ways this is possible if the white flowers are to be separated in any arrangement (Flowers of
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Number of triangles =
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∴
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Solution.
The first vertex can be selected in 20 ways. The remaining two are to be selected from 17 vertices so
that they are not consecutive. This can be done in 17C2 – 16 ways.
∴
The total number of ways = 20 × (17C2 – 16)
But in this method, each selection is repeated thrice.
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same colour are identical).
Solution.
Total we have 12 flowers 3 red, 4 yellow and 5 white.
12 !
Number of arrangements = 3 ! 4 ! 5 ! = 27720.
For the second part, first arrange 3 red & 4 yellow
7!
This can be done in 3 ! 4 ! = 35 ways
4!
The consonants in their positions can be arranged in 2 ! = 12 ways.
3!
The vowels in their positions can be arranged in 2 ! = 3 ways
∴
Total number of arrangements = 12 × 3 = 26
Self Practice Problems :
13.
How many words can be formed using the letters of the word ASSESSMENT if each word begin with A
and end with T.
Ans. 840
14.
If all the letters of the word ARRANGE are arranged in all possible ways, in how many of words we will
have the A's not together and also the R's not together.
Ans. 660
15.
How many arrangements can be made by taking four letters of the word MISSISSIPPI.
Ans. 176.
6.
Formation of Groups :
Number of ways in which (m + n + p) different things can be divided into three different groups containing
m, n & p things respectively is
(m + n + p) !
m!n! p!
,
If m = n = p and the groups have identical qualitative characteristic then the number of groups =
(3n)!
.
n! n! n! 3!
However, if 3n things are to be divided equally among three people then the number of ways =
(3n)!
(n!)3
.
Example # 14
12 different toys are to be distributed to three children equally. In how many ways this can be done.
Solution.
The problem is to divide 12 different things into three different groups.
Number of ways =
Example # 15
12 !
= 34650.
4! 4! 4!
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Example # 13
In how many ways the letters of the word "ARRANGE" can be arranged without altering the relative
positions of vowels & consonants.
Solution.
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Now select 5 places from among 8 places (including extremes) & put the white flowers there.
This can be done in 8C5 = 56.
∴
The number of ways for the 2nd part = 35 × 56 = 1960.
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In how many ways 10 persons can be divided into 5 pairs.
Solution.
We have each group having 2 persons and the qualitative characteristic are same (Since there is no
purpose mentioned or names for each pair).
10 !
Thus the number of ways =
= 945.
( 2 ! )5 5 !
9 persons enter a lift from ground floor of a building which stops in 10 floors (excluding ground floor). If
is known that persons will leave the lift in groups of 2, 3, & 4 in different floors. In how many ways this
can happen.
Ans. 907200
17.
In how many ways one can make four equal heaps using a pack of 52 playing cards.
52 !
Ans.
(13 ! ) 4 4 !
18.
In how many ways 11 different books can be parcelled into four packets so that three of the packets
contain 3 books each and one of 2 books, if all packets have the same destination.
11 !
Ans.
(3 ! ) 4 2
7.
Selection of one or more objects
(a)
(b)
(c)
Number of ways in which atleast one object be selected out of 'n' distinct objects is
n
C1 + nC2 + nC3 +...............+ nCn = 2n – 1
Number of ways in which atleast one object may be selected out of 'p' alike objects of one type
'q' alike objects of second type and 'r' alike of third type is
(p + 1) (q + 1) (r + 1) – 1
Number of ways in which atleast one object may be selected from 'n' objects where 'p' alike of
one type 'q' alike of second type and 'r' alike of third type and rest
n – (p + q + r) are different, is
(p + 1) (q + 1) (r + 1) 2n – (p + q + r) – 1
Example # 16
There are 12 different books on a shelf. In how many ways we can select atleast one of them.
Solution.
We may select 1 book, 2 books,........, 12 books.
∴
The number of ways = 12C1 + 12C2 + ....... + 12C12 = 212 – 1. = 4095
Example # 17
There are 12 fruits in a basket of which 5 are apples, 4 mangoes and 3 bananas (fruits of same species
are identical). How many ways are there to select atleast one fruit.
Solution.
Let x be the number of apples being selected
y be the number of mangoes being selected and
z be the number of bananas being selected.
Then x = 0, 1, 2, 3, 4, 5
y = 0, 1, 2, 3, 4
z = 0, 1, 2, 3
Total number of triplets (x, y, z) is 6 × 5 × 4 = 120
Exclude (0, 0, 0)
∴
Number of combinations = 120 – 1 = 119.
Self Practice Problems
19.
In a shelf there are 5 physics, 4 chemistry and 3 mathematics books. How many combinations are
there if (i) books of same subject are different (ii) books of same subject are identical.
Ans. (i) 4095
(ii) 119
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16.
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Self Practice Problems :
From 5 apples, 4 mangoes & 3 bananas in how many ways we can select atleast two fruits of each
variety if (i) fruits of same species are identical (ii) fruits of same species are different.
Ans. (i)
24
(ii)
1144
8.
Multinomial Theorem:
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20.
For example the number of ways in which a selection of four letters can be made from the
letters of the word PROPORTION is given by coefficient of x4 in
(1 + x + x2 + x3) (1 + x + x2) (1 + x + x2) (1 + x) (1 + x) (1 + x).
(ii)
Method of fictious partition :
Number of ways in which n identical things may be distributed among p persons if each person
may receive none, one or more things is; n+p−1Cn.
Example # 18
Find the number of solutions of the equation x + y + z = 6, where x, y, z ∈ W.
Solution.
Number of solutions
= coefficient of x6 in (1 + x + x2 + ....... x6)3
= coefficient of x6 in (1 – x7)3 (1 – x) –3
= coefficient of x6 in (1 – x)–3
 3 + 6 − 1 8
 = C6 = 28.
= 
6


Example # 19
In a bakery four types of biscuits are available. In how many ways a person can buy 10 biscuits if he
decide to take atleast one biscuit of each variety.
Solution.
Let x be the number of biscuits the person select from first variety, y from the second, z from the third
and w from the fourth variety. Then the number of ways = number of solutions of the equation
x + y + z + w = 10.
where x = 1, 2, .........,7
y = 1, 2, .........,7
z = 1, 2, .........,7
w = 1, 2, .........,7
This is equal to = coefficient of x10 in (x + x2 + ...... + x7)4
= coefficient of x6 in (1 + x + ....... + x6)4
= coefficient of x6 in (1 – x7)4 (1 – x)–4
= coefficient x6 in (1 – x)–4
 4 + 6 − 1
 = 84.
= 
6


Self Practice Problems:
21.
Three distinguishable dice are rolled. In how many ways we can get a total 15.
Ans. 10.
22.
In how many ways we can give 5 apples, 4 mangoes and 3 oranges (fruits of same species are similar)
to three persons if each may receive none, one or more.
Ans. 3150
9.
Let N = pa. qb. rc...... where p, q, r...... are distinct primes & a, b, c..... are natural numbers then :
(a)
(b)
The total numbers of divisors of N including 1 & N is = (a + 1) (b + 1) (c + 1)........
The sum of these divisors is =
(p0 + p1 + p 2 +.... + pa) (q0 + q1 + q2 +.... + q b) (r0 + r 1 + r2 +.... + r c)........
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(i)
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Coefficient of xr in expansion of (1 − x)−n = n+r−1Cr (n ∈ N)
Number of ways in which it is possible to make a selection from m + n + p = N things, where p are alike
of one kind, m alike of second kind & n alike of third kind taken r at a time is given by coefficient of
xr in the expansion of
(1 + x + x2 +...... + xp) (1 + x + x2 +...... + x m) (1 + x + x2 +...... + x n).
=
1
2
+ 1)....
[(a + 1)(b + 1)(c + 1)....+1]
if N is not a perfect square
if N is a perfect square
Number of ways in which a composite number N can be resolved into two factors which are
relatively prime (or coprime) to each other is equal to 2 n−1 where n is the number of different
prime factors in N.
Example # 20
Find the number of divisors of 1350. Also find the sum of all divisors.
Solution.
1350 = 2 × 33 × 52
∴
Number of divisors = (1+ 1) (3 + 1) (2 + 1) = 24
sum of divisors = (1 + 2) (1 + 3 + 32 + 33) (1 + 5 + 5 2) = 3720.
Example # 21
In how many ways 8100 can be resolved into product of two factors.
Solution.
8100 = 22 × 34 × 52
Number of ways =
1
((2 + 1) (4 + 1) (2 + 1) + 1) = 23
2
Self Practice Problems :
23.
How many divisors of 9000 are even but not divisible by 4. Also find the sum of all such divisors.
Ans. 12, 4056.
24.
In how many ways the number 8100 can be written as product of two coprime factors.
Ans. 4
10.
Let there be 'n' types of objects, with each type containing atleast r objects. Then the number of ways
of arranging r objects in a row is nr.
Example # 22
How many 3 digit numbers can be formed by using the digits 0, 1, 2, 3, 4, 5. In how many of these we
have atleast one digit repeated.
Solution.
We have to fill three places using 6 objects (repeatation allowed), 0 cannot be at 100th place. The
number of numbers = 180.
Number of numbers in which no digit is repeated = 100
∴
Number of numbers in which atleast one digit is repeated = 180 – 100 = 80
Example # 23
How many functions can be defined from a set A containing 5 elements to a set B having 3 elements.
How many these are surjective functions.
Solution.
Image of each element of A can be taken in 3 ways.
∴
Number of functions from A to B = 35 = 243.
Number of into functions from A to B = 25 + 25 + 2 5 – 3 = 93.
∴
Number of onto functions = 150.
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(d)
1 (a + 1)(b + 1)(c
2
9 of 27
Number of ways in which N can be resolved as a product of two factors is
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(c)
10 of 27
Find the sum of all three digit numbers those can be formed by using the digits. 0, 1, 2, 3, 4.
Ans. 27200.
26.
How many functions can be defined from a set A containing 4 elements to a set B containing 5 elements.
How many of these are injective functions.
Ans. 625, 120
27.
In how many ways 5 persons can enter into a auditorium having 4 entries.
Ans. 1024.
11. Dearrangement :
Number of ways in which 'n' letters can be put in 'n' corresponding envelopes such that no letter goes
to correct envelope is

1 1 1 1
n 1

n ! 1 − + − + .......... .. + ( −1)
n! 
 1! 2! 3 ! 4 !
Example # 24
In how many ways we can put 5 writings into 5 corresponding envelopes so that no writing go to the
corresponding envelope.
Solution.
The problem is the number of dearragements of 5 digits.
 1
1
1
1
+
−  = 44.
This is equal to 5!  −
 2 ! 3 ! 4 ! 5! 
Example # 25
Four slip of papers with the numbers 1, 2, 3, 4 written on them are put in a box. They are drawn one by
one (without replacement) at random. In how many ways it can happen that the ordinal number of
atleast one slip coincide with its own number.
Solution.
Total number of ways = 4 ! = 24.
The number of ways in which ordinal number of any slip does not coincide with its own number is the
 1
1
1
number of dearrangements of 4 objects = 4 !  2 ! − 3 ! + 4 !  = 9


Thus the required number of ways. = 24 – 9 = 15
Self Practice Problems:
28.
In a match column question, Column Ι contain 10 questions and Column II contain 10 answers written
in some arbitrary order. In how many ways a student can answer this question so that exactly 6 of his
matchings are correct.
Ans. 1890
29.
In how many ways we can put 5 letters into 5 corresponding envelopes so that atleast one letter go to
wrong envelope.
Ans. 119
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25.
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Self Practice Problems :
DEFINITIONS :
1.
PERMUTATION : Each of the arrangements in a definite order which can be made by taking some or
all of a number of things is called a PERMUTATION.
2.
11 of 27
Shor
evision
Shortt R
Re
COMBINATION : Each of the groups or selections which can be made by taking some or all of a
number of things without reference to the order of the things in each group is called a COMBINATION.
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
If nPr denotes the number of permutations of n different things, taking r at a time, then
n!
nP = n (n − 1) (n − 2)..... (n − r + 1) =
Note that , nPn = n !.
r
( n − r )!
If nCr denotes the number of combinations of n different things taken r at a time, then
n
n!
Pr
nC =
=
where r ≤ n ; n ∈ N and r ∈ W.
r
r!(n − r )!
r!
The number of ways in which (m + n) different things can be divided into two groups containing m & n
(m + n ) !
things respectively is :
If m = n, the groups are equal & in this case the number of subdivision
m!n!
is ( 2n )! ; for in any one way it is possible to interchange the two groups without obtaining a new
n! n!2!
distribution. However, if 2n things are to be divided equally between two persons then the number of
(2n )!
ways =
.
n!n!
Number of ways in which (m + n + p) different things can be divided into three groups containing m , n
( m + n + p )!
, m ≠ n ≠ p.
& p things respectively is
m! n! p!
(3n )!
If m = n = p then the number of groups =
.
n!n!n!3!
(3n )!
However, if 3n things are to be divided equally among three people then the number of ways =
.
( n!) 3
The number of permutations of n things taken all at a time when p of them are similar & of one type, q of
them are similar & of another type, r of them are similar & of a third type & the remaining
n – (p + q + r) are all different is : n! .
p!q!r!
The number of circular permutations of n different things taken all at a time is ; (n − 1)!. If clockwise &
anti−clockwise circular permutations are considered to be same, then it is
(viii)
( n −1)!
.
2
Note : Number of circular permutations of n things when p alike and the rest different taken all at a time
( n − 1)!
distinguishing clockwise and anticlockwise arrangement is
.
p!
Given n different objects, the number of ways of selecting atleast one of them is ,
nC + nC + nC +.....+ nC = 2n − 1. This can also be stated as the total number of combinations of n
1
2
3
n
distinct things.
(ix)
Total number of ways in which it is possible to make a selection by taking some or all out of
p + q + r +...... things , where p are alike of one kind, q alike of a second kind , r alike of third kind &
so on is given by :
(p + 1) (q + 1) (r + 1)........ –1.
(x)
Number of ways in which it is possible to make a selection of m + n + p = N things , where p are alike
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RESULTS :
(i)
A Useful Notation : n! = n (n − 1) (n − 2)......... 3. 2. 1 ; n ! = n. (n − 1) !
0! = 1! = 1 ; (2n)! = 2n. n ! [1. 3. 5. 7...(2n − 1)]
Note that factorials of negative integers are not defined.
98930 58881
FUNDAMENTAL PRINCIPLE OF COUNTING :
If an event can occur in ‘m’ different ways, following which another event can occur in ‘n’ different ways,
then the total number of different ways of simultaneous occurrence of both events in a definite order is
m × n. This can be extended to any number of events.
(xi)
Number of ways in which n distinct things can be distributed to p persons if there is no restriction to the
number of things received by men = pn.
(xii)
Number of ways in which n identical things may be distributed among p persons if each person may
receive none , one or more things is ; n+p−1Cn.
(xiii)
a.
nC
r
= nCn−r ; nC0 = nCn = 1
c.
nC
r
+ nCr−1 =
nC
x
= nCy ⇒ x = y or x + y = n
r
nC is maximum if : (a) r = n if n is even. (b) r = n −1 or n +1 if n is odd.
r
2
2
2
Let N = pa. qb. rc...... where p , q , r...... are distinct primes & a , b , c..... are natural numbers then:
(a)
The total numbers of divisors of N including 1 & N is = (a + 1)(b + 1)(c + 1).....
(b)
The sum of these divisors is
= (p0 + p1 + p2 +.... + pa) (q0 + q1 + q2 +.... + qb) (r0 + r1 + r2 +.... + rc)....
(c)
Number of ways in which N can be resolved as a product of two
factors is =
1 (a + 1)(b + 1)(c + 1)....
2
1
2
[(a + 1)(b + 1)(c + 1).... + 1]
if N is not a perfect square
if N is a perfect square
(d)
(xvi)
Number of ways in which a composite number N can be resolved into two factors which are
relatively prime (or coprime) to each other is equal to 2n−1 where n is the number of different
prime factors in N.
[ Refer Q.No.28 of Ex−I ]
Grid Problems and tree diagrams.
DEARRANGEMENT :
Number of ways in which n letters can be placed in n directed letters so that no letter goes into its own
1 1 1
n 1 
envelope is = n!  − + +...........+(−1)
.
n! 
 2! 3! 4!
(xvii) Some times students find it difficult to decide whether a problem is on permutation or combination or
both. Based on certain words / phrases occuring in the problem we can fairly decide its nature as per the
following table :
PROBLEMS OF COMBINATIONS PROBLEMS OF PERMUTATIONS
" Selections , choose
" Distributed group is formed
" Committee
" Geometrical problems
" Arrangements
" Standing in a line seated in a row
" problems on digits
" Problems on letters from a word
EXER
CISE–1
EXERCISE–1
Q.1
The straight lines l1 , l2 & l3 are parallel & lie in the same plane. A total of m points are taken on the line
l1 , n points on l2 & k points on l3. How many maximum number of triangles are there whose vertices are
at these points ?
Q.2
How many five digits numbers divisible by 3 can be formed using the digits 0, 1, 2, 3, 4, 7 and 8 if each
digit is to be used atmost once.
Q.3
There are 2 women participating in a chess tournament. Every participant played 2 games with the other
participants. The number of games that the men played between themselves exceeded by 66 as compared
to the number of games that the men played with the women. Find the number of participants & the total
numbers of games played in the tournament.
Q.4
All the 7 digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once, and not divisible by
5 are arranged in the increasing order. Find the (2004)th number in this list.
Q.5
5 boys & 4 girls sit in a straight line. Find the number of ways in which they can be seated if 2 girls are
98930 58881
(xv)
b.
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(xiv)
n+1C
;
12 of 27
of one kind , m alike of second kind & n alike of third kind taken r at a time is given by coefficient of xr
in the expansion of
(1 + x + x2 +...... + xp) (1 + x + x2 +...... + xm) (1 + x + x2 +...... + xn).
Note : Remember that coefficient of xr in (1 − x)−n = n+r−1Cr (n ∈ N). For example the number of ways
in which a selection of four letters can be made from the letters of the word PROPORTION is given by
coefficient of x4 in (1 + x + x2 + x3) (1 + x + x2) (1 + x + x2) (1 + x) (1 + x) (1 + x).
Q.7
An examination paper consists of 12 questions divided into parts A & B.
Part-A contains 7 questions & Part−B contains 5 questions. A candidate is required to attempt 8 questions
selecting atleast 3 from each part. In how many maximum ways can the candidate select the questions ?
Q.8
In how many ways can a team of 6 horses be selected out of a stud of 16 , so that there shall always be
3 out of A B C A ′ B ′ C ′ , but never A A ′ , B B ′ or C C ′ together.
Q.9
During a draw of lottery, tickets bearing numbers 1, 2, 3,......, 40, 6 tickets are drawn out & then
arranged in the descending order of their numbers. In how many ways, it is possible to have 4th ticket
bearing number 25.
Q.10
Find the number of distinct natural numbers upto a maximum of 4 digits and divisible by 5, which can be
formed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 each digit not occuring more than once in each number.
Q.11
The Indian cricket team with eleven players, the team manager, the physiotherapist and two umpires are to
travel from the hotel where they are staying to the stadium where the test match is to be played. Four of them
residing in the same town own cars, each a four seater which they will drive themselves. The bus which was to
pick them up failed to arrive in time after leaving the opposite team at the stadium. In how many ways can they
be seated in the cars ? In how many ways can they travel by these cars so as to reach in time, if the seating
arrangement in each car is immaterial and all the cars reach the stadium by the same route.
Q.12
There are n straight lines in a plane, no 2 of which parallel , & no 3 pass through the same point. Their
point of intersection are joined. Show that the number of fresh lines thus introduced is
n ( n − 1)(n − 2)( n − 3)
.
8
In how many ways can you divide a pack of 52 cards equally among 4 players. In how many ways the
cards can be divided in 4 sets, 3 of them having 17 cards each & the 4th with 1 card.
Q.13
Q.14
A firm of Chartered Accountants in Bombay has to send 10 clerks to 5 different companies, two clerks
in each. Two of the companies are in Bombay and the others are outside. Two of the clerks prefer to
work in Bombay while three others prefer to work outside. In how many ways can the assignment be
made if the preferences are to be satisfied.
Q.15
A train going from Cambridge to London stops at nine intermediate stations. 6 persons enter the train
during the journey with 6 different tickets of the same class. How many different sets of ticket may they
have had?
Q.16 Prove that if each of m points in one straight line be joined to each of n in another by straight lines
1
terminated by the points, then excluding the given points, the lines will intersect mn(m – 1)(n –1) times.
4
Q.17 How many arrangements each consisting of 2 vowels & 2 consonants can be made out of the letters of
the word ‘DEVASTATION’?
Q.18
Find the number of words each consisting of 3 consonants & 3 vowels that can be formed from the
letters of the word “Circumference”. In how many of these c’s will be together.
Q.19
There are 5 white , 4 yellow , 3 green , 2 blue & 1 red ball. The balls are all identical except for colour.
These are to be arranged in a line in 5 places. Find the number of distinct arrangements.
Q.20
How many 4 digit numbers are there which contains not more than 2 different digits?
Q.21
(a)
(b)
(c)
In how many ways 8 persons can be seated on a round table
If two of them (say A and B) must not sit in adjacent seats.
If 4 of the persons are men and 4 ladies and if no two men are to be in adjacent seats.
If 8 persons constitute 4 married couples and if no husband and wife, as well as no two men, are to be
in adjacent seats?
Q.22
(i)
If 'n' things are arranged in circular order , then show that the number of ways of selecting four of
the things no two of which are consecutive is
n ( n − 5) ( n − 6 ) ( n − 7 )
4!
(ii)
If the 'n' things are arranged in a row, then show that the number of such sets of four is
13 of 27
A crew of an eight oar boat has to be chosen out of 11 men five of whom can row on stroke side only, four
on the bow side only, and the remaining two on either side. How many different selections can be made?
98930 58881
Q.6
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together & the other 2 are also together but separate from the first 2.
Q.24
How many ten digits whole number satisfy the following property they have 2 and 5 as digits, and there
are no consecutive 2's in the number (i.e. any two 2's are separated by at least one 5).
Q.25
How many different ways can 15 Candy bars be distributed between Ram, Shyam, Ghanshyam and
Balram, if Ram can not have more than 5 candy bars and Shyam must have at least two. Assume all
Candy bars to be alike.
Q.26
Find the number of distinct throws which can be thrown with 'n' six faced normal dice which are
indistinguishable among themselves.
Q.27
How many integers between 1000 and 9999 have exactly one pair of equal digit such as 4049 or 9902
but not 4449 or 4040?
Q.28
In a certain town the streets are arranged like the lines of a chess board. There are 6 streets running north
& south and 10 running east & west. Find the number of ways in which a man can go from the north-west
corner to the south-east corner covering the shortest possible distance in each case.
Q.29
(i)
(ii)
(iii)
(iv)
(v)
Q.30
There are 20 books on Algebra & Calculus in our library. Prove that the greatest number of selections
each of which consists of 5 books on each topic is possible only when there are 10 books on each topic
in the library.
Prove that : nPr = n−1Pr + r. n−1Pr−1
If 20Cr+2 = 20C2r−3 find 12Cr
Find the ratio 20Cp to 25Cr when each of them has the greatest value possible.
Prove that n−1C3 + n−1C4 > nC3 if n > 7.
Find r if 15C3r = 15Cr+3
EXER
CISE–2
EXERCISE–2
Q.1
Q.2
Find the number of ways in which 3 distinct numbers can be selected from the set
{31, 32, 33, ....... 3100, 3101} so that they form a G.P.
Let n & k be positive integers such that n ≥ k(k+1) . Find the number of solutions
2
(x1 , x2 ,.... , xk) , x1 ≥ 1 , x2 ≥ 2 ,... , xk ≥ k , all integers, satisfying x1 + x2 + .... + xk = n.
Q.3
There are counters available in 7 different colours. Counters are all alike except for the colour and they
are atleast ten of each colour. Find the number of ways in which an arrangement of 10 counters can be
made. How many of these will have counters of each colour.
Q.4
For each positive integer k, let Sk denote the increasing arithmetic sequence of integers whose first term
is 1 and whose common difference is k. For example, S3 is the sequence 1, 4, 7, 10...... Find the number
of values of k for which Sk contain the term 361.
Q.5
Find the number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed
using the letters of the word "DIFFERENTIATION".
Q.6
A shop sells 6 different flavours of ice-cream. In how many ways can a customer choose 4 ice-cream
cones if
(i)
they are all of different flavours
(ii)
they are non necessarily of different flavours
(iii)
they contain only 3 different flavours
(iv)
they contain only 2 or 3 different flavours?
Q.7
6 white & 6 black balls of the same size are distributed among 10 different urns. Balls are alike except
for the colour & each urn can hold any number of balls. Find the number of different distribution of the
balls so that there is atleast 1 ball in each urn.
Q.8
There are 2n guests at a dinner party. Supposing that the master and mistress of the house have fixed
seats opposite one another, and that there are two specified guests who must not be placed next to one
14 of 27
(a)
(b)
(c)
98930 58881
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( n − 3) ( n − 4) ( n − 5) ( n − 6)
4!
How many divisors are there of the number x = 21600. Find also the sum of these divisors.
In how many ways the number 7056 can be resolved as a product of 2 factors.
Find the number of ways in which the number 300300 can be split into 2 factors which are
relatively prime.
Q.10
How many 15 letter arrangements of 5 A's, 5 B's and 5 C's have no A's in the first 5 letters, no B's in the
next 5 letters, and no C's in the last 5 letters.
Q.11
5 balls are to be placed in 3 boxes. Each box can hold all 5 balls. In how many different ways can we
place the balls so that no box remains empty if,
(i) balls & boxes are different
(ii) balls are identical but boxes are different
(iii) balls are different but boxes are identical (iv) balls as well as boxes are identical
(v) balls as well as boxes are identical but boxes are kept in a row.
Q.12
(i)
(ii)
(iii)
In how many other ways can the letters of the word MULTIPLE be arranged;
without changing the order of the vowels
keeping the position of each vowel fixed &
without changing the relative order/position of vowels & consonants.
Q.13
Find the number of ways in which the number 30 can be partitioned into three unequal parts, each part
being a natural number. What this number would be if equal parts are also included.
In an election for the managing committee of a reputed club , the number of candidates contesting
elections exceeds the number of members to be elected by r (r > 0). If a voter can vote in 967 different
ways to elect the managing committee by voting atleast 1 of them & can vote in 55 different ways to elect
(r − 1) candidates by voting in the same manner. Find the number of candidates contesting the elections
& the number of candidates losing the elections.
Q.14
Q.15
Find the number of three digits numbers from 100 to 999 inclusive which have any one digit that is the
average of the other two.
Q.16
Prove by combinatorial argument that :
n+1C = nC + nC
(a)
r
r
r–1
n + mc = nc · mc + nc · mc
n
m
n
m
(b)
r
0
r
1
r − 1 + c2 · cr − 2 +....... + cr · c0.
Q.17
A man has 3 friends. In how many ways he can invite one friend everyday for dinner on 6 successive
nights so that no friend is invited more than 3 times.
Q.18
12 persons are to be seated at a square table, three on each side. 2 persons wish to sit on the north side
and two wish to sit on the east side. One other person insists on occupying the middle seat (which may
be on any side). Find the number of ways they can be seated.
Q.19
There are 15 rowing clubs; two of the clubs have each 3 boats on the river; five others have each 2 and
the remaining eight have each 1; find the number of ways in which a list can be formed of the order of the
24 boats, observing that the second boat of a club cannot be above the first and the third above the
second. How many ways are there in which a boat of the club having single boat on the river is at the
third place in the list formed above?
Q.20
25 passengers arrive at a railway station & proceed to the neighbouring village. At the station there are
2 coaches accommodating 4 each & 3 carts accommodating 3 each. Find the number of ways in which
they can proceed to the village assuming that the conveyances are always fully occupied & that the
conveyances are all distinguishable from each other.
Q.21
An 8 oared boat is to be manned by a crew chosen from 14 men of which 4 can only steer but can not
row & the rest can row but cannot steer. Of those who can row, 2 can row on the bow side. In how
many ways can the crew be arranged.
Q.22
How many 6 digits odd numbers greater than 60,0000 can be formed from the digits 5, 6, 7, 8, 9, 0 if
(i) repetitions are not allowed
(ii) repetitions are allowed.
Q.23
Find the sum of all numbers greater than 10000 formed by using the digits 0 , 1 , 2 , 4 , 5 no digit being
repeated in any number.
Q.24
The members of a chess club took part in a round robin competition in which each plays every one else
once. All members scored the same number of points, except four juniors whose total score were 17.5.
How many members were there in the club? Assume that for each win a player scores 1 point , for draw
1/2 point and zero for losing.
15 of 27
Each of 3 committees has 1 vacancy which is to be filled from a group of 6 people. Find the number of
ways the 3 vacancies can be filled if ;
(i)
Each person can serve on atmost 1 committee.
(ii)
There is no restriction on the number of committees on which a person can serve.
(iii)
Each person can serve on atmost 2 committees.
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another. Show that the number of ways in which the company can be placed is (2n − 2)!.(4n2 − 6n + 4).
Q.26
Six faces of an ordinary cubical die marked with alphabets A, B, C, D, E and F is thrown n times and the
list of n alphabets showing up are noted. Find the total number of ways in which among the alphabets
A, B, C, D, E and F only three of them appear in the list.
Q.27
Find the number of integer betwen 1 and 10000 with at least one 8 and at least one 9 as digits.
Q.28
The number of combinations n together of 3n letters of which n are 'a' and n are 'b' and the rest unlike is
(n + 2). 2n − 1.
Q.29
In Indo−Pak one day International cricket match at Sharjah , India needs 14 runs to win just before the
start of the final over. Find the number of ways in which India just manages to win the match (i.e. scores
exactly 14 runs) , assuming that all the runs are made off the bat & the batsman can not score more than
4 runs off any ball.
Q.30
A man goes in for an examination in which there are 4 papers with a maximum of m marks for each
paper; show that the number of ways of getting 2m marks on the whole is
1
(m + 1) (2m² + 4m + 3).
3
EXER
CISE–3
EXERCISE–3
Q.1
Find the total number of ways of selecting five letters from the letters of the word INDEPENDENT.
[ REE '97, 6 ]
Q.2
(i)
Select the correct alternative(s).
Number of divisors of the form 4n + 2 ( n ≥ 0) of the integer 240 is
(A) 4
(B) 8
(C) 10
(ii)
Q.3
Q.4
Q.5
Q.6
[ JEE ’98, 2 + 2 ]
(D) 3
An n-digit number is a positive number with exactly 'n' digits. Nine hundred distinct n-digit numbers are
to be formed using only the three digits 2, 5 & 7. The smallest value of n for which this is possible is :
(A) 6
(B) 7
(C) 8
(D) 9
How many different nine digit numbers can be formed from the number 223355888 by rearranging its
digits so that the odd digits occupy even positions ?
[JEE '2000, (Scr)]
(A) 16
(B) 36
(C) 60
(D) 180
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of
[ JEE '2001, (Scr) ]
' n ' sides. If Tn + 1 − Tn = 21 , then ' n ' equals:
(A) 5
(B) 7
(C) 6
(D) 4
The number of arrangements of the letters of the word BANANA in which the two N’s do not appear
adjacently is
[JEE 2002 (Screening), 3]
(A) 40
(B) 60
(C) 80
(D) 100
Number of points with integral co-ordinates that lie inside a triangle whose co-ordinates are
(0, 0), (0, 21) and (21,0)
[JEE 2003 (Screening), 3]
(A) 210
(B) 190
(C) 220
(D) None
(n 2 ) !
Q.7
Using permutation or otherwise, prove that
is an integer, where n is a positive integer.
( n!) n
[JEE 2004, 2 out of 60]
Q.8
A rectangle with sides 2m – 1 and 2n – 1 is divided into squares of unit length
by drawing parallel lines as shown in the diagram, then the number of rectangles
possible with odd side lengths is
(B) 4m + n – 1
(A) (m + n + 1)2
2 2
(C) m n
(D) mn(m + 1)(n + 1)
[JEE 2005 (Screening), 3]
Q.9
If r, s, t are prime numbers and p, q are the positive integers such that their LCM of p, q is is r2t4s2, then
the numbers of ordered pair of (p, q) is
(A) 252
(B) 254
(C) 225
(D) 224 [JEE 2006, 3]
16 of 27
98930 58881
(a)
(b)
There are 3 cars of different make available to transport 3 girsls and 5 boys on a field trip. Each car can
hold up to 3 children. Find
the number of ways in which they can be accomodated.
the numbers of ways in which they can be accomodated if 2 or 3 girls are assigned to one of the cars.
In both the cars internal arrangement of childrent inside the car is to be considered as immaterial.
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Q.25
17 of 27
EXER
CISE–4
EXERCISE–4
There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number
of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the
same colour, is:
(A) 6 (7 ! − 4 !)
(B) 7 (6 ! − 4 !)
(C) 8 ! − 5 !
(D) none
2.
The number of permutations that can be formed by arranging all the letters of the word ‘NINETEEN’ in
which no two E’s occur together is
5!
8!
5!
8!
(B)
(C) 3 ! × 6C3
(D) 5 ! × 6C3.
(A) 3! 3!
6
3! × C 2
3.
The number of ways in which n different things can be given to r persons when there is no restriction as
to the number of things each may receive is:
(A) nCr
(B) nPr
(C) nr
(D) rn
4.
The number of divisors of apbqcrds where a, b, c, d are primes & p, q, r, s ∈ N, excluding 1 and the
number itself is:
(A) p q r s
(B) (p + 1) (q + 1) (r + 1) (s + 1) − 4
(C) p q r s − 2
(D) (p + 1) (q + 1) (r + 1) (s + 1) − 2
5.
The number of ordered triplets of positive integers which are solutions of the equation x + y + z = 100
is:
(A) 3125
(B) 5081
(C) 6005
(D) 4851
6.
Number of ways in which 7 people can occupy six seats, 3 seats on each side in a first class railway
compartment if two specified persons are to be always included and occupy adjacent seats on the
same side, is (k). 5 ! then k has the value equal to:
(A) 2
(B) 4
(C) 8
(D) none
7.
Number of different words that can be formed using all the letters of the word "DEEPMALA" if two
vowels are together and the other two are also together but separated from the first two is:
(A) 960
(B) 1200
(C) 2160
(D) 1440
8.
Six persons A, B, C, D, E and F are to be seated at a circular table. The number of ways this can be
done if A must have either B or C on his right and B must have either C or D on his right is:
(A) 36
(B) 12
(C) 24
(D) 18
9.
The number of ways in which 15 apples & 10 oranges can be distributed among three persons, each
receiving none, one or more is:
(A) 5670
(B) 7200
(C) 8976
(D) none of these
10.
The number of permutations which can be formed out of the letters of the word "SERIES" taking three
letters together is:
(A) 120
(B) 60
(C) 42
(D) none
11.
Seven different coins are to be divided amongst three persons. If no two of the persons receive the
same number of coins but each receives atleast one coin & none is left over, then the number of ways
in which the division may be made is:
(A) 420
(B) 630
(C) 710
(D) none
12.
The streets of a city are arranged like the lines of a chess board. There are m streets running North to
South & 'n' streets running East to West. The number of ways in which a man can travel from NW to SE
corner going the shortest possible distance is:
(A)
13.
m2 + n 2
(m − 1)2 . (n − 1)2
(C)
( m + n) !
m! . n!
(D)
(m + n − 2) !
(m − 1) ! . (n − 1) !
In a conference 10 speakers are present. If S1 wants to speak before S 2 & S2 wants to speak after
S3, then the number of ways all the 10 speakers can give their speeches with the above restriction if the
remaining seven speakers have no objection to speak at any number is:
(A) 10C3
14.
(B)
(B) 10P8
(C) 10P3
(D)
10 !
3
Two variants of a test paper are distributed among 12 students. Number of ways of seating of the
students in two rows so that the students sitting side by side do not have identical papers & those
sitting in the same column have the same paper is:
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98930 58881
Part : (A) Only one correct option
(12)!
2 5 . 6!
(C) (6 !)2. 2
(D) 12 ! × 2
15.
Sum of all the numbers that can be formed using all the digits 2, 3, 3, 4, 4, 4 is:
(A) 22222200
(B) 11111100
(C) 55555500
(D) 20333280
16.
There are m apples and n oranges to be placed in a line such that the two extreme fruits being both
oranges. Let P denotes the number of arrangements if the fruits of the same species are different and
Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the
value equal to:
(B) mP2. nPn. (n − 2) !
(C) nP2. nPn. (m − 2) !
(D) none
(A) nP2. mPm. (n − 2) !
17.
The number of integers which lie between 1 and 106 and which have the sum of the digits equal to 12 is:
(A) 8550
(B) 5382
(C) 6062
(D) 8055
18.
Number of ways in which a pack of 52 playing cards be distributed equally among four players so that
each may have the Ace, King, Queen and Jack of the same suit is:
(A)
36 !
(B)
(9 !)
4
36 ! . 4 !
(9 !)
4
(C)
36 !
(9 !)4
. 4!
(D) none
19.
A five letter word is to be formed such that the letters appearing in the odd numbered positions are
taken from the letters which appear without repetition in the word "MATHEMATICS". Further the letters
appearing in the even numbered positions are taken from the letters which appear with repetition in the
same word "MATHEMATICS". The number of ways in which the five letter word can be formed is:
(A) 720
(B) 540
(C) 360
(D) none
20.
Number of ways of selecting 5 coins from coins three each of Rs. 1, Rs. 2 and Rs. 5 if coins of the
same denomination are alike, is:
(A) 9
(B) 12
(C) 21
(D) none
21.
Number of ways in which all the letters of the word " ALASKA " can be arranged in a circle distinguishing
between the clockwise and anticlockwise arrangement, is:
(A) 60
(B) 40
(C) 20
(D) none of these
22.
If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r2 t4s2, then the
number of ordered pair (p, q) is
[IIT – 2006]
(A) 252
(B) 254
(C) 225
(D) 224
Part : (B) May have more than one options correct
23.
C6 + nC4 > n + 2C5 − nC5 for all ' n ' greater than:
(A) 8
(B) 9
(C) 10
n+1
(D) 11
24.
In an examination, a candidate is required to pass in all the four subjects he is studying. The number
of ways in which he can fail is
(A) 4P1 + 4P2 + 4P3 + 4P4
(B) 44 – 1
4
(D) 4C1 + 4C2 + 4C3 + 4C4
(C) 2 – 1
25.
The kindergarten teacher has 25 kids in her class. She takes 5 of them at a time, to zoological garden
as often as she can, without taking the same 5 kids more than once. Then the number of visits, the
teacher makes to the garden exceeds that of a kid by:
(A) 25C5 − 24C4
(B) 24C5
(C) 25C5 − 24C5
(D) 24C4
The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if the letter C are
separated from one another is:
26.
12 !
(A) 13C3. 5 ! 3 ! 2!
27.
28.
13 !
(B) 5 ! 3 ! 3 ! 2 !
13 !
(D) 11. 6 !
There are 10 points P1, P2,...., P10 in a plane, no three of which are collinear. Number of straight lines
which can be determined by these points which do not pass through the points P1 or P2 is:
(A) 10C2 − 2. 9C1
(B) 27
(C) 8C2
(D) 10C2 − 2. 9C1 + 1
Number of quadrilaterals which can be constructed by joining the vertices of a convex polygon of
20 sides if none of the side of the polygon is also the side of the quadrilateral is:
15
(A) C4 − C2
17
29.
14 !
(C) 3 ! 3 ! 2 !
15
(B)
C 3 . 20
4
(C) 2275
(D) 2125
You are given 8 balls of different colour (black, white,...). The number of ways in which these balls can
be arranged in a row so that the two balls of particular colour (say red & white) may never come
together is:
(A) 8 ! − 2.7 !
(B) 6. 7 !
(C) 2. 6 !. 7C2
(D) none
18 of 27
(B)
98930 58881
12!
6! 6!
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(A)
A man is dealt a poker hand (consisting of 5 cards) from an ordinary pack of 52 playing cards. The
number of ways in which he can be dealt a "straight" (a straight is five consecutive values not of the
same suit, eg. {Ace, 2 , 3, 4 , 5}, {2, 3, 4, 5, 6}.......................... & {10 , J , Q , K, Ace}) is
(B) 4 !. 210
(C) 10. 210
(D) 10200
(A) 10 (4 5 − 4)
31.
Number of ways in which 3 numbers in A.P. can be selected from 1, 2, 3,...... n is:
32.
(n −1)2
4
if n is odd
(D)
n (n − 2)
4
n (n − 2)
4
if n is odd
if n is even
Consider the expansion , (a 1 + a2 + a 3 +....... + ap)n where n ∈ N and n ≤ p. The correct statement(s) is/
are:
(A)
number of different terms in the expansion is, n + p − 1C n
(B)
co-efficient of any term in which none of the variables a1, a2 ..., ap occur more than once is ' n '
(C)
co-efficient of any term in which none of the variables a1, a2,..., ap occur more than once is n ! if
n=p
(D)
 p
Number of terms in which none of the variables a1, a2,......, ap occur more than once is   .
 n
EXER
CISE–5
EXERCISE–5
1.
In a telegraph communication how many words can be communicated by using atmost 5 symbols.
(only dot and dash are used as symbols)
2.
If all the letters of the word 'AGAIN' are arranged in all possible ways & put in dictionary order, what is
the 50th word.
3.
A committee of 6 is to be chosen from 10 persons with the condition that if a particular person 'A' is
chosen, then another particular person B must be chosen.
4.
A family consists of a grandfather, m sons and daughters and 2n grand children. They are to be seated
in a row for dinner. The grand children wish to occupy the n seats at each end and the grandfather
refuses to have a grand children on either side of him. In how many ways can the family be made to sit?
5.
The sides AB, BC & CA of a triangle ABC have 3, 4 & 5 interior points respectively on them. Find the
number of triangles that can be constructed using these interior points as vertices.
6.
How many five digits numbers divisible by 3 can be formed using the digits 0, 1, 2, 3, 4, 7 and 8 if, each
digit is to be used atmost one.
7.
In how many other ways can the letters of the word MULTIPLE be arranged ; (i) without changing the
order of the vowels (ii) keeping the position of each vowel fixed (iii) without changing the relative order/
position of vowels & consonants.
8.
There are p intermediate stations on a railway line from one terminus to another. In how many ways can
a train stop at 3 of these intermediate stations if no 2 of these stopping stations are to be consecutive?
9.
Find the number of positive integral solutions of x + y + z + w = 20 under the following conditions:
(i)
Zero values of x, y, z, w are include
(ii)
Zero values are excluded
(iii)
No variable may exceed 10; Zero values excluded
(iv)
Each variable is an odd number
(v)
x, y, z, w have different values (zero excluded).
10.
Find the number of words each consisting of 3 consonants & 3 vowels that can be formed from the
letters of the word “CIRCUMFERENCE”. In how many of these C’s will be together.
11.
If ' n ' distinct things are arranged in a circle, show that the number of ways of selecting three of these
things so that no two of them are next to each other is,
12.
1
n (n − 4) (n − 5).
6
In maths paper there is a question on "Match the column" in which column A contains 6 entries & each
entry of column A corresponds to exactly one of the 6 entries given in column B written randomly.
2 marks are awarded for each correct matching & 1 mark is deducted from each incorrect matching.
98930 58881
(C)
(B)
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 n − 1
(A) 
 if n is even
 2 
19 of 27
30.
13.
Show that the number of combinations of n letters together out of 3n letters of which n are a and n are
b and the rest unlike is, (n + 2). 2n − 1.
14.
Find the number of positive integral solutions of, (i) x2 − y2 = 352706 (ii) xyz = 21600
15.
There are ' n ' straight line in a plane, no two of which are parallel and no three pass through the same
point. Their points of intersection are joined. Show that the number of fresh lines thus introduced is,
20 of 27
A student having no subjective knowledge decides to match all the 6 entries randomly. Find the number
of ways in which he can answer, to get atleast 25 % marks in this question.
A forecast is to be made of the results of five cricket matches, each of which can be a win or a draw or
a loss for Indian team. Find
(i)
number of forecasts with exactly 1 error
(ii)
number of forecasts with exactly 3 errors
(iii)
number of forecasts with all five errors
17.
Prove by permutation or otherwise
18.
 n + 1
 (2 n+1 – n – 2) where n > 1, and the rund scored
If total number of runs scored in n matches is 
 4 
in the kth match are given by k. 2n+1–k, where 1 ≤ k ≤ n. Find n
[IIT – 2005]
(n )!
2
(n !)n
is an integer (n ∈ I+).
[IIT – 2004]
ANSWER KEY
EXER
CISE–1
EXERCISE–1
− (mC3 + nC3 + kC3)
Q.1
m+n+kC
Q.4
4316527
Q.5
43200
Q.8
960
Q.9
24C
Q.13
52!
(13! ) 4
3
;
52 !
3!(17 ! ) 3
Q.18 22100 , 52
2
.
15C
3
Q.2
744
Q.3
13 , 156
Q.6
145
Q.7
420
Q.11
12! ;
Q.10 1106
45 C
Q.14 5400
Q.15
Q.19 2111
Q.20 576
6
11! . 4!
(3!) 4 2!
Q.17 1638
Q.21 (a) 5 · (6!) , (b) 3! · 4!, (c) 12
Q.23
(a) 72 ; 78120 ; (b) 23 ; (c) 32
Q.24 143
Q.25 440
Q.26
n + 5C
5
Q.29
143
; (v) r = 3
4025
Q.28
(14 )!
5!9!
(ii) 792 ; (iii)
Q.27 3888
EXER
CISE–2
EXERCISE–2
where m = (1/2) (2n − k² + k − 2)
Q.1
2500
Q.2
mC
Q.3
 49 
710 ;   10 !
 6
Q.4
24
Q.5
532770
Q.6
(i) 15, (ii) 126, (iii) 60, (iv) 105
Q.9
120, 216, 210
Q.10 2252
Q.12
(i) 3359 ; (ii) 59 ; (iii) 359
Q.15 121
24!
23!
8C .
Q.19
2
5 ;
1 (3!) 2 (2!) 5
(3!) (2!)
k−1
Q.11
Q.7
26250
(i) 150 ; (ii) 6 ; (iii) 25 ; (iv) 2 ; (v) 6
Q.13 61, 75
Q.14 10, 3
Q.17 510
Q.18 2 ! 3 ! 8 !
Q.20
(25) !
(3!) (4!) 4 . 4
3
Q.21 4 . (4!)² . 8C4 . 6C2
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16.
98930 58881
1
8 n (n − 1) (n − 2) (n − 3).
Q.23 3119976
Q.25 (a) 1680; (b) 1140
Q.26
Q.27 974
Q.29 1506
6C
3[3
n
21 of 27
Q.22 240 , 15552
Q.24 27
– 3C1(2n – 2) – 3C2]
72
Q.2
(i) A ; (ii) B
Q.5
A
Q.6
B
1. A
2. C
3. D
4. D
5. D
6. C
7. D
8. D
9. C
10. C
11. B
12. D
13. D
14. D
15. A
16. A
17. C
18. B
19. B
20. B
21. C
22. C
23. BCD
Q.3
Q.8
C
C
EXER
CISE–4
EXERCISE–4
24. CD
27. CD 28. AB 29. ABC
25. AB
26. AD
30. AD 31. CD 32. ACD
EXER
CISE–5
EXERCISE–5
1. 62
2. NAAIG
4. (2n)! m! (m − 1)
7. (i) 3359
8.
p–2
5. 205
3. 154
6. 744
(ii) 59 (iii) 359
C3
9. (i) 23C3 (ii) 19C3 (iii) 19C3 – 4.9C3 (iv) 11C8 (v) 552
10. 22100, 52 12. 56 ways
14. (i) Zero (ii) 1260
18. 7
16. (i) 10
(ii) 80 (iii) 32
Q.4
B
Q.9
C
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Q.1
98930 58881
EXER
CISE–3
EXERCISE–3