## Combinatorics

 Question 1

The number of permutations of the characters in LILAC so that no character appears in its original position, if the two L’s are indistinguishable, is _______.

 A 12
Question 1 Explanation:
There are 5 places.
― ― ― ― ―
Given: L I L A C
The derangement formula ⎣n!/e⎦ cannot be directly performed as there are repeated characters.
Let’s proceed in regular manner:
The L, L can be placed in other ‘3’ places as

(1) Can be arranged such that A, I, C be placed in three positions excluding ‘C’ being placed at its own position, which we get only 2×2×1 = 4 ways.
Similarly (2) can be filled as A, I, C being placed such that 4th position is not filled by A, so we have 2×2×1 = 4 ways. Similarly with (3).
Totally, we get 4+4+4 = 12 ways.
 Question 2

Two girls have picked 10 roses, 15 sunflowers and 15 daffodils. What is the number of ways they can divide the flowers amongst themselves?

 A 1638 B 2100 C 2640 D None of the above
Question 2 Explanation:
Formula for distributing n identical objects into r persons is,
n+r-1Cr-1
So for 10 roses,
10+2-1C2-1 = 11C1 = 11
For 15 sunflowers,
15+2-1C2-1 = 16C1 = 16
For 15 daffodils,
15+2-1C2-1 = 16C1 = 16
11×16×16 = 2816
 Question 3

The  number  of  binary  strings  of  n  zeroes  and  k  ones  that  no  two  ones  are adjacent is

 A n-1Ck B nCk C nCk+1 D None of the above
Question 3 Explanation:
Since there are n zeroes, so
XOXOXOXOXOXOX
n+1 gaps can be possible, where 1's can be placed so that no two one's are adjacent. So, no. of ways in which k 1's can be placed in n+1 gaps are,
n+1Ck
 Question 4

The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52 cards to guarantee that three cards are from some same suit is

 A 3 B 8 C 9 D 12
Question 4 Explanation:
No. of cards = 52
No. of suits = 4(P)
Apply pigeon hole principal.
Then number of pigeons = n
floor [(n-1)/P] + 1 = 3
floor [(n-1)/P] = 2
floor [(n-1)] = 8
floor (n) = 8 + 1
n ≥ 9
Minimum no. of cards, n = 9
 Question 5

How many 4-digit even numbers have all 4 digits distinct?

 A 2240 B 2296 C 2620 D 4536
Question 5 Explanation:
If the last digit is 0 then

If last digit is (2, 4, 6, 8)

Total possibilities = 504 + 1792 = 2296
 Question 6

(a) In how many ways can a given positive integer n ≥ 2 be expressed as the sum of 2 positive integers (which are not necessarily distinct). For example, for n = 3, the number of ways is 2, i.e., 1+2, 2+1. Give only the answer without any explanation.

(b) In how many ways can a given positive integer n ≥ 3 be expressed as the sum of 3 positive integers (which are not necessarily distinct). For example, for n = 4, the number of ways is 3, i.e., 1+2+1, 2+1+1. Give only the answer without any explanation.

(c) In how many ways can a given positive integer n ≥ k be expressed as the sum of k positive integers (which are not necessarily distinct)? Give only the answer without explanation.

 A Theory Explanation is given below.
 Question 7

m identical balls are to be placed in n distinct bags. You are given that m ≥ kn, where, k is a natural number ≥1. In how many ways can the balls be placed in the bags if each bag must contain at least k balls?

 A B C D
Question 7 Explanation:
Since we want at least k balls in each bag, so first we put kn balls into bags, k balls in each bag. Now we are left with m - kn balls, and we have to put them into n bags such that each bag may receive 0 or more balls. So applying theorem 2 of stars and bars with m - nk stars and n bars, we get number of ways to be

. So option (B) is correct.
 Question 8

n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is

 A B C D
Question 8 Explanation:
The possibilities to attend party is
i) Both husband and wife comes
ii) Only wife comes
iii) Both are not come
The no. of different gatherings possible at party is
= 3 * 3 * 3 * 3 * ... n times
= 3n
 Question 9

Consider the sequence <xn>, n>= 0 defined by the recurrence relation xn + 1 = c . xn2- 2, where c > 0.
Suppose there exists a non-empty, open interval (a, b) such that for all x0 satisfying a < x0 < b, the sequence converges to a limit. The sequence converges to the value?

 A B C 2 D
Question 9 Explanation:
Let c=1, x0=1
Then,
x1 = c ⋅ (x0)2 - 2 = 1 ⋅ (1)2 - 2 = -1
x2 = c ⋅ (x1)2 - 2 = 1 ⋅ (-1)2 - 2 = -1
So, the value converges to -1, which is equal to

Exactly, only (B) is answer. As all the term of x converges to -1.
 Question 10

Consider the sequence <xn>, n>= 0 defined by the recurrence relation xn + 1 = c. xn2- 2, where c > 0.
For which of the following values of c, does there exist a non-empty open interval (a,b) such that the sequence xn, converges for all x0 satisfying a < x0

(i) 0.25
(ii) 0.35
(iii) 0.45
(iv) 0.5
 A (i) Only B (i) and (ii) only C (i), (ii) and (iii) only D (i), (ii), (iii) and (iv)
Question 10 Explanation:
For the series to converge the limit: 'n' tends to infinity of (xn+1 / xn) should be <1.
From the recurrence we should have c(xn)2 - xn - 2 < 0
For all the above values of c we have above equation as negative.
 Question 11

The exponent of 11 in the prime factorization of 300! is

 A 27 B 28 C 29 D 30
Question 11 Explanation:
 Question 12

In how many ways can b blue balls and r red balls be distributed in n distinct boxes?

 A B C D
Question 12 Explanation:
No. of blue ball distributed to n boxes in
C(n+b-1, b) = (n+b-1)!/ (n-1)!b!
No. of red distributed to n boxes is
(n+r-1, r) = (n+r-1)!/ (n-1)!r!
Total no. of ways = (n+b-1)!/(n-1)!b! * (n+r-1)!/(n-1)!r!
= (n+b-1)!(n+r-1)!/(n-1)!b!(n-1)!r!
 Question 13
Consider solving the following system of simultaneous equations using LU decomposition.
 A B C D
Question 13 Explanation:
The LU decomposition of the given matrix is

Given,

(Assume x1,x2,x3 as a,b,c)

 Question 14
Which one of the following is the decimal form of generating function of the sequences {an} n >=0 defined below
 A B C D
Question 14 Explanation:

 Question 15

Choose the correct alternatives (More than one may be correct).

The number of ways in which 5 A's, 5 B's and 5 C's can be arranged in a row is:
 A 15!/(5!)3 B 15! C (15/5) D 15!(5!3!)
Question 15 Explanation:
The no. of ways,
 Question 16

Mala has a colouring book in which each English letter is drawn two times. She wants to paint each of these 52 prints with one of k colours, such that the colour-pairs used to colour any two letters are different. Both prints of a letter can also be coloured with the same colour. What is the minimum value of k that satisfies this requirement?

 A 9 B 8 C 7 D 6
Question 16 Explanation:
No. of letters from A-Z is = 26
Each is printed twice the no. of letters = 26×2 = 52
If Mala has k colours, she can have k pairs of same colours.
She also can have kC2 different pairs in which each pair is having different colours.
So total no. of pairs that can be coloured = k+kC2
k+kC2 ≥ 26
k+k(k-1)/2 ≥ 26
k(k+1)/2 ≥ 26
k(k+1) ≥ 52
k(k+1) ≥ 7*8
k≥7
 Question 17

The number of divisors of 2100 is ______.

 A 36 B 37 C 38 D 39
Question 17 Explanation:
Let N = 2100
= 22+3×52×7 (i.e., product of primes)
Then the number of divisions of 2100 is
(2+1)∙(1+1)∙(2+1)∙(1+1) i.e., (3)(2)(3)(2) i.e., 36.
 Question 18
The value of the definite integral

is___(Rounded off to the nearest integer)
 A 0
 Question 19

Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1. Let a99 = K × 104. The value of K is ___________.

 A 198 B 199 C 200 D 201
Question 19 Explanation:
an = 6n2 + 2n + a(n-1)
Replace a(n-1)

⇒ an = 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯ a1
Given that a1 = 8, replace it
⇒ an = 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯8
= 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯ + 6(1)2 + 2(1)

= 6(n2 + (n-1)2 + (n-2)2 + ⋯ + 22 + 12) + 2(n + (n-1) + ⋯1)
Sum of n2 = (n(n+1)(2n+1))/6
Sum of n = (n(n+1))/2
= 6 × (n(n+1)(2n+1))/6 + 2×(n(n+1))/2
= n(n+1)[1+2n+1]
= n(n+1)[2n+2]
= 2n(n+1)2
Given a99 = k×104
a99 = 2(99)(100)2 = 198 × 104
∴k = 198
 Question 20

The coefﬁcient of x12 in (x3 + x4 + x5 + x6 + ...)3 is _________.

 A 10 B 11 C 12 D 13
Question 20 Explanation:
Co-efficient of x12 in (x3 + x4 + x5 + x6+...)3
⇒ [x3(1 + x + x2 + x3 + ...)]3
= x9(1 + x + x2 + x3 + ...)3
First Reduction:
As x9 is out of the series, we need to find the co-efficient of x3 in (1 + x + x2 + ⋯)3

Here, m=3, k=3, the coefficient

= 5C3 = 5!/2!3! = 10
 Question 21

Let an be the number of n-bit strings that do NOT contain two consecutive 1s. Which one of the following is the recurrence relation for an?

 A an = a(n-1) + 2a(n-2) B an = a(n-1) + a(n-2) C an = 2a(n-1) + a(n-2) D an = 2a(n-1) + 2a(n-2)
Question 21 Explanation:
an = number of n-bit strings, that do not have two consecutive 1’s.
If n=1, we have {0,1}
# Occurrences = 2
If n=2, we have {00,01,10}
# Occurrences = 3
If n=3, we have {000,001,010,100,101}
# Occurrences = 5
It is evident that a3 = a1 + a2
Similarly, an = an-1 + an-2
 Question 22

If the ordinary generating function of a sequence is , then a3 - a0 is equal to ________.

 A 15 B 16 C 17 D 18
Question 22 Explanation:
 Question 23

The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 is _________.

 A 271 B 272 C 273 D 274
Question 23 Explanation:

Let A = number divisible by 3
B = numbers divisible by 5
C = number divisible by 7
We need to find “The number of integers between 1 and 500 that are divisible by 3 or 5 or 7" i.e., |A∪B∪C|
We know,
|A∪B∪C| = |A|+|B|+C-|A∩B|-|A∩C|-|B∩C|+|A∩B|
|A| = number of integers divisible by 3
[500/3 = 166.6 ≈ 166 = 166]
|B| = 100
[500/5 = 100]
|C| = 71
[500/7 = 71.42]
|A∩B| = number of integers divisible by both 3 and 5 we need to compute with LCM (15)
i.e.,⌊500/15⌋ ≈ 33
|A∩B| = 33
|A∩C| = 500/LCM(3,7) 500/21 = 23.8 ≈ 28
|B∩C| = 500/LCM(5,3) = 500/35 = 14.48 ≈ 14
|A∩B∩C| = 500/LCM(3,5,7) = 500/163 = 4.76 ≈ 4
|A∪B∪C| = |A|+|B|+|C|-|A∩B|-|A∩C|-|B∩C|+|A∩B∩C|
= 166+100+71-33-28-14+4
= 271
 Question 24

Which one of the following is a closed form expression for the generating function of the sequence {an}, where an = 2n+3 for all n = 0, 1, 2, …?

 A 3/(1-x)2 B 3x/(1-x)2 C 2-x/(1-x)2 D 3-x/(1-x)2
Question 24 Explanation:
 Question 25

The number of 4 digit numbers having their digits in non-decreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3} is _____.

 A 15 B 16 C 17 D 18
Question 25 Explanation:
Just try to write all the four digit numbers containing the digits 1, 2 and 3 only, in decreasing order.
1 1 1 1
1 1 1 2
1 1 1 3
1 1 2 2
1 1 2 3
1 1 3 3
1 2 2 2
1 2 2 3
1 2 3 3
1 2 3 3
1 3 3 3
2 2 2 2
2 2 2 3
2 2 3 3
2 3 3 3
3 3 3 3
Hence, total 15 4-digit no. are possible.
 Question 26

A pennant is a sequence of numbers, each number being 1 or 2. An n-pennant is a sequence of numbers with sum equal to n. For example, (1,1,2) is a 4-pennant. The set of all possible 1-pennants is {(1)}, the set of all possible 2-pennants is {(2), (1,1)}and the set of all 3-pennants is {(2,1), (1,1,1), (1,2)}. Note that the pennant (1,2) is not the same as the pennant (2,1). The number of 10-pennants is ________.

 A 89 B 90 C 91 D 92
Question 26 Explanation:
No twos: 1111111111 ⇒ 1 pennant
Single two: 211111111 ⇒ 9!/8!1! = 9 pennants
Two twos: 22111111 ⇒ 8!/6!2! = 28
Three twos: 2221111 ⇒ 7!/3!4! = 35
Four twos: 222211 ⇒ 6!/4!2! = 15
Five twos: 22222 ⇒ 1
Total = 89 pennants.
 Question 27

Let , where |x|<1. What is g(i)?

 A i B i+1 C 2i D 2i
Question 27 Explanation:

Put g(i) = i+1

S = 1 + 2x + 3x2 + 4x3 + .....
Sx = 1x + 2x2 + 3x3 + 4x4 + ......
S - Sx = 1 + x + x2 + x3 + .....
[Sum of infinite series in GP with ratio < 1 is a/1-r]
S - Sx = 1/(1-x)
S(1-x) = 1/(1-x)
S = 1/(1-x)2
 Question 28

Given a set of elements N = {1, 2, ..., n} and two arbitrary subsets A⊆N and B⊆N, how many of the n! permutations π from N to N satisfy min(π(A)) = min(π(B)), where min(S) is the smallest integer in the set of integers S, and π(S) is the set of integers obtained by applying permutation π to each element of S?

 A (n-|A ∪ B|) |A| |B| B (|A|2+|B|2)n2 C n!(|A∩B|/|A∪B|) D
Question 28 Explanation:
Given a set of elements N = {1, 2, 3, ...N}
Two arbitrary subsets A⊆N and B⊆N.
Out of n! permutations π from N to N, to satisfy
min(π(A)) = min (π(B))
*) π(S) is the set of integers obtained by applying permutation π to each element of S.
If min(π(A)) =min (π(B)), say y = π(x) is the common minimum.
Since the permutation π is a 1-to-1 mapping of N,
x ∈ A∩B
∴ A∩B cannot be empty.
⇒ y = π(x)
= π(A∩B) is the minimum of π(A∪B) is the minimum of π(A) and π(B) are to be same.
You can think like
*) If the minimum of π(A) and π(B) are same [min π(A)] = min [π(B)]
then min(π(A∩B)) = min(π(A∪B))
∴ Total number is given by n! |A∩B|/|A∪B|
*) Finally
Considering all possible permutations, the fraction of them that meet this condition |π(A∩B)| / |π(A∪B)|
[The probability of single permutation].
Ex: N = {1, 2, 3, 4} A = {1, 3} B = {1, 2, 4}

Since π is one to one mapping
|π(A∩B)| = |A∩B|
∴ π(A) = {1, 2}
π(B) = {1, 4, 3}
π(A∩B) = {1}
π(A∪B) = {1, 2, 3, 4}
4! × 1/4 = 6
 Question 29

The number of distinct positive integral factors of 2014 is _________.

 A 0.26 B 0.27 C 8 D 0.29
Question 29 Explanation:
First lets find prime factorization of 2014
= 2' × 19' × 53'
Now number of distinct integral factors of 2014 will be,
(1+1)×(1+1)×(1+1) = 2×2×2 = 8
 Question 30

Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i,j) then it can move to either (i+1,j) or (i,j+1).

How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0,0)?

 A B 220 C 210 D None of the above
Question 30 Explanation:
Lets name one unit up movement as 'u' and one unit right movement as 'r'.
So now we have 10 u's and 10 r's, i.e.,
uuuuuuuuuurrrrrrrrrr
So, finally the no. of arrangements of above sequences is,
There are 30 questions to complete.

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