Combinatorics

Question 1
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
       Engineering-Mathematics       Combinatorics       Gate 2018
Question 1 Explanation: 
Question 2
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
       Engineering-Mathematics       Combinatorics       Gate 2017 set-01
Question 2 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 3
A
15
B
16
C
17
D
18
       Engineering-Mathematics       Combinatorics       GATE 2017(set-02)
Question 3 Explanation: 
Question 4
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)
       Engineering-Mathematics       Combinatorics       2016 set-01
Question 4 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 5
The coefficient of x12 in (x+ x+ x+ x+ ...)3 is ___________.
A
10
B
11
C
12
D
13
       Engineering-Mathematics       Combinatorics       2016 set-01
Question 5 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 6
Consider the recurrence relation a= 8, a= 6n+ 2n + an-1. Let a99 = K × 104. The value of K is ___________.
A
198
B
199
C
200
D
201
       Engineering-Mathematics       Combinatorics       2016 set-01
Question 6 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 7
The number of divisors of 2100 is ______.  
A
36
B
37
C
38
D
39
       Engineering-Mathematics       Combinatorics       GATE 2015 -(Set-2)
Question 7 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 8
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
       Engineering-Mathematics       Combinatorics       GATE 2015(Set-03)
Question 8 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 9
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
       Engineering-Mathematics       Combinatorics       GATE 2014(Set-01)
Question 9 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 10
The number of distinct positive integral factors of 2014 is _________
A
0.26
B
0.27
C
8
D
0.29
       Engineering-Mathematics       Combinatorics       Gate 2014 Set -02
Question 10 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 11
The exponent of 11 in the prime factorization of 300! is
A
27
B
28
C
29
D
30
       Engineering-Mathematics       Combinatorics       Gate 2008-IT
Question 11 Explanation: 
300/11 = 27
27/11 = 2
2/11 = 0
-------------------
29
-------------------
The exponent of 11 in the prime factorization of 300! = 29
Question 12
In how many ways can b blue balls and r red balls be distributed in n distinct boxes?
A
B
C
D
       Engineering-Mathematics       Combinatorics       Gate 2008-IT
Question 12 Explanation: 
Question 13
 
A
B
220
C
210
D
None of the above
       Engineering-Mathematics       Combinatorics       Gate-2007
Question 13 Explanation: 
At each move, robot can move either 1 unit right, or 1 unit up, and there will be 20 such moves required to reach (10,10) from (0,0). So we have to divide these 20 moves, numbered from 1 to 20, into 2 groups : right group and up group. Right group contains those moves in which we move right, and up group contains those moves in which we move up. Each group contains 10 elements each. So basically, we have to divide 20 things into 2 groups of 10 10 things each, which can be done in 20! / (10!∗10! )=20C10 ways.
Question 14

A
29
B
219
C
D
       Engineering-Mathematics       Combinatorics       Gate-2007
Question 14 Explanation: 
Since we are not allowed to traverse from (4,4) to (5,4). So we will subtract all those paths which were passing through (4,4) to (5,4).
To count number of paths passing through (4,4) to (5,4), we find number of paths from (0,0) to (4,4), and then from (5,4) to (10,10).
From (0,0) to (4,4), number of paths = 8C4 (found in same way as in previous question).
From (5,4) to (10,10), number of paths = 11C5.
So total number of paths required : 20C108C411C5
Question 15
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
       Engineering-Mathematics       Combinatorics       Gate-2006
Question 15 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 16
     
A
i
B
i+1
C
2i
D
2i
       Engineering-Mathematics       Combinatorics       Gate-2005
Question 16 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 17

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
       Engineering-Mathematics       Combinatorics       Gate-2004
Question 17 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 18
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
       Engineering-Mathematics       Combinatorics       Gate-2003
Question 18 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 19
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
       Engineering-Mathematics       Combinatorics       Gate-2003
Question 19 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 20
 
A
Theory Explanation is given below.
       Engineering-Mathematics       Combinatorics       Gate-2002
Question 21
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
       Engineering-Mathematics       Combinatorics       Gate-2000
Question 21 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 22
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
       Engineering-Mathematics       Combinatorics       Gate-1999
Question 22 Explanation: 

If n zeros are present then n+1 gaps are possible. So we have to fill (n+1) number of k's.
So, total possible ways is (n+1)Ck.
Question 23
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
       Engineering-Mathematics       Combinatorics       Gate-1999
Question 23 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
∴ The final answer is,
11×16×16 = 2816
Question 24
How many 4-digit even numbers have all 4 digits distinct?
A
2240
B
2296
C
2620
D
4536
       Engineering-Mathematics       Combinatorics       Gate-2001
Question 24 Explanation: 
If the last digit is 0 then

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

Total possibilities = 504 + 1792 = 2296
There are 24 questions to complete.