Q1: Let U = {1, 2,...,n}, where n is a large positive integer greater than 1000. Let k be a positive integer less than n. Let A, B be subsets of U with |A| = |B| = k and A ∩ B = ϕ. We say that a permutation of U separates A from B if one of the following is true.
- All members of A appear in the permutation before any of the members of B.
- All members of B appear in the permutation before any of the members of A.
How many permutations of U separate A from B? (2023)
(a) n!
(b) 
(c)
(d)
Ans: (d)
Sol: |A| = |B| = k
elements of A followed by elements of B ⇒ k!k!
elements of B followed by elements of A ⇒ k!k!
1 → number of permutations of 2k elements of A and B = 2k!k! ways
this will create 2k + 1 positions for the rest of n - 2k elements.
p objects can be placed in q places such that each place can have 0 to p objects in C(p + q - 1, q - 1) ways
2 → there for n - 2k elements can be placed in 2k+1 positions in C(n-2k+2k+1-1, 2k+1-1) = C(n, 2k) ways.
3 → Also these n-2k elements can also be permuted ⇒ (n - 2k)!
So total number of permutations of U which separates A from B = 2(k!)(k!)C(n, 2k)(n-2k)!, which is option D.
Q2: There are 6 jobs with distinct difficulty levels, and 3 computers with distinct processing speeds. Each job is assigned to a computer such that:
The fastest computer gets the toughest job and the slowest computer gets the easiest job.
Every computer gets at least one job.
The number of ways in which this can be done is ___________. (2021 SET-1)
(a) 18
(b) 36
(c) 65
(d) 27
Ans: (c)
Sol: Let C1 be the fastest and C3 be the slowest computers.
These two are assigned two jobs. Now out of the remaining 4 jobs we need to ensure C2 gets at least 1. Without this constraint we can assign 4 jobs to 3 computers in 34 = 81 ways. Out of these 81 ways 24 = 16 will be having no jobs for C2.
So, number of possible ways so that C2 gets at least one job = 81 - 16 = 65.
Q3: 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 ______. (2020)
(a) 120
(b) 24
(c) 36
(d) 12
Ans: (d)
Sol: We have LILAC.
Lets number the positions as 1, 2, 3, 4, 5. Now the two Ls cannot be placed at position 1 and 3 but they can be positioned at 2, 4, 5 in 3C2 = 3 ways. (since the Ls are indistinguishable)
Now one of 2, 4, 5 is vacant. Without loss of generality lets say 2 is vacant.
Now, if 2 is vacant we can't palce I there but we can place any of A, or C, so we have 2 choices for the position which is left after filling the two Ls. Now all of 2, 4, 5 are filled.
For the remaining two places 1, 3 we have two characters left and none of them is L so we can place them in 2! = 2 ways.
Multiply them all = 3C2 ∗ 2 ∗ 2! = 3 ∗ 2 ∗ 2 = 12 (ans).
Q4: The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 is ______. (2017 SET-1)
(a) 135
(b) 350
(c) 271
(d) 335
Ans: (c)
Sol: Here, we can apply the property of set. Let Dn denote divisibility by n, Dn1,n2 denote divisibility by both n1 and n2 and so on.
N(D3∪D5∪D7) = N(D3) + N(D5) + N(D7) − N(D3,5) − N(D5,7) − N(D3,7) + N(D3,5,7)
= 166 + 100 + 71 − 33 − 14 − 23 + 4
= 271
Q5: 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____________. (2015 SET-3)
(a) 10
(b) 15
(c) 18
(d) 12
Ans: (b)
Sol: We can arrive at a solution by constructing a graph for each starting digit. For example root 3 means - starting with 3 it can have 3 children 1, 2, 3 and the construction goes.
3 can have three children 1, 2, 3
2 can have two children 1, 2
1 can have only 1 as child.
Graph need to be done till four levels as we need 4 digits and we have 3 such graphs starting with 3, 2 and 1.
And finally count the total number of leaves of all the graphs gives our answer as 15.
Q6: The number of divisors of 2100 is _____ . (2015 SET-2)
(a) 28
(b) 34
(c) 36
(d) 40
Ans: (c)
Sol: 2100 = 7 × 3 × 22 × 52
Hence, total number of factors will be = (1 + 1) × (1 + 1) × (2 + 1) × (2 + 1) = 2 × 2 × 3 × 3 = 36,
because any factor is obtained by multiplying the prime factors zero or more times. (one extra for zero)
Q7: The number of bit strings of length 8 that will either start with 1 or end with 00 is? (2014)
(a) 32
(b) 128
(c) 160
(d) 192
Ans: (c)
Sol: String starting with 1 - 8 places 1 place fixed .7 places have 2 choices . 27 = 128
ending with 00- 8 places 2 fixed. = 26 =64
Common strings will be there that have been counted twice are . starting with 1 and ending with 00. such number of string will be .
Fix 3 position rest have 2 choices = 32
Total = 128 + 64 - 32 = 160
Q8: The number of distinct positive integral factors of 2014 is ____ (2014 SET-2)
(a) 4
(b) 8
(c) 9
(d) 10
Ans: (b)
Sol: First do prime factorization of 2014−21 × 191 × 531
Now to get a factor of 2014, we can choose any combination of the prime factors including 0. i.e; 20 and 21 are possible and similarly for other prime factors also, there are 2 possibilities. So, the total number of positive integral factors = 2 × 2 × 2 = 8
(When all the powers of prime factors are 0, we get 1 and when all the powers are maximum, we get the given number.)
Q9: 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 _____. (2014 SET-1)
(a) 1024
(b) 89
(c) 156
(d) 112
Ans: (b)
Sol: Let us denote number of n−pennants by f(n), so f(10) is number of 10-pennants.
A 10−pennant means sum of numbers in sequence is 10. If we look at any 9−pennant, we can make it a 10−pennant by adding 1 into that sequence. Similarly, we can make any 8−pennant a 10−pennant by adding 2 into that sequence.
So all 10−pennants can be formed by 8−pennants and 9−pennants, and no other pennant (since we can add only 1 or 2 into a sequence)
So f(10) = f(9) + f(8)
This is in fact a Fibonacci sequence, in which f(1) = 1, f(2) = 2, so this sequence becomes 1, 2, 3, 5, 8, 13, 21, 34, 55, 89…
So f(10) = 89.
Q10: There are 5 bags labelled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Then the product of the labels of the bags having 11 gm coins is ___. (2014 SET-1)
(a) 12
(b) 10
(c) 16
(d) 8
Ans: (a)
Sol: Suppose X is the number of coins of 11 gm and Y is the number of 10 gm coins.
According to question,
11X + 10Y = 323 → (1)
X + Y = 31 → (2)
Solving (1), (2), we get X = 13 and Y = 18. So, here number of coins of 11 gm is 13 and the only possible combination for 13 coins is
- 1 coin from bag1
- 4 coins from bag3
- 8 coins from bag4
So, product of label of bags will be = 1 x 3 x 4 = 12.
