## Number of even divisions of 504 is

Q: Number of even divisions of 504 is

(A) 12

(B) 24

(C) 6 3C2

(D) 18

Sol. 504 = 23 × 32 ×7.. Any even divisor of 504 is of the form 2i × 3i × 7k, where I ≤ i ≤ 3 , 0 ≤ j ≤2, 0 ≤ k ≤ 1. Thus total number of even divisors is
3 × 3 ×2 = 18
Hence (C), (D) are the correct answer.

## The number of ways in which a mixed double game can be arranged amongst 9 married couples if no husband and wife play in the same game is

Q: The number of ways in which a mixed double game can be arranged amongst 9 married couples if no husband and wife play in the same game is

(A) 756

(B) 1512

(C) 2 . 9C27C2

(D) none of these

Sol. We can choose two men out of 9 in 9C2 ways. Since no husband and wife are to play in the same game, two women out of the remaining 7 can be chosen in 7C2 ways.

If M1, M2, W1 and W2 are chosen, then a team may consist of M1 and W1 or M1 and W2. Thus the number of ways of arranging the game is
(9C2) (7C2)(2) = 1512.

Hence (B), (C) are the correct answer.

## Let A be the set of 4-digit numbers a1 a2 a3 a4 where a1> a2> a3> a4, then n(A) is equal to

Q: Let A be the set of 4-digit numbers a1 a2 a3 a4 where a1> a2> a3> a4, then n(A) is equal to

(A) 126

(B) 10C4

(C) 210

(D) none of these

Sol. Any selection of four digits from the ten digits 0, 1, 2, 3, . . . , 9 gives one such number.

So, the required number of numbers = 10C4 = 210 .

Hence (B), (C) are the correct answer.

## Triplet (x, y, z) is chosen from the set {1, 2, 3,…. n }, such that x ≤ y < z . The number of such triplets is

Q: Triplet (x, y, z) is chosen from the set {1, 2, 3,…. n }, such that x ≤ y < z . The number of such triplets is

(A)$\frac{n(n-1)(n+1)}{2}$

(B) nC3

(C) nC2

(D) nC2 + nC3

Sol. Any three numbers x, y, z from {1, 2, 3, . . . .} can be chosen in nC3 ways and we get unique triplet ( x, y, z) , x< y < z .

Again any two numbers x, z can be chosen from {1, 2, 3, . . . , n } in nC2 ways and we get the triplet ( x, x, z) , x< z .

Hence total number of required triplets is nC2 + nC3

Hence (A), (D) are the correct answer.

## A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the elements of P…

Q: A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the elements of P. A subset Q of A is again chosen. The number of ways of chosen P and Q so that P ∩ Q = φ is

(A) 22n2nCn

(B) 2n

(C) 2n – 1

(D) 3n

Sol. Let A = { a1, a2 , a3 , . . . , an} . For ai ∈ A, we have the following choices:

(i) ai  ∈ P and ai ∉ Q

(ii) ai ∈ P and ai ∉ Q

(iii) ai ∉ P and ai ∈ Q

(iv) ai ∉ P and ai ∉ Q

Out of these only (ii) , (iii) and (iv) imply ai  ∉ P ∩ Q. Therefore, the number of ways in which none of a1, a2, . . .an belong to P ∩ Q is 3n.

Hence (D) is the correct answer.

## The number of solutions of x1+x2+x3 = 51 (x1,x2,x3 being odd natural numbers) is

Q: The number of solutions of x1 + x2 + x3 = 51 (x1 ,x2 , x3 being odd natural numbers) is

(A) 300

(B) 325

(C) 330

(D) 350

Sol. Let odd natural numbers be 2a – 1 , 2b -1 , 2c – 1 ; where a, b, c are natural numbers

2a – 1 + 2b- 1 + 2c- 1 = 51

⇒ a + b + c = 27

a ≥ 1, b ≥ 1, c ≥ 1 …(1)

No. of solutions of (1) is coefficient of x24 in (1-x)-3

= 26C2 = 13 × 25 = 325

Hence (B) is the correct answer.

## The number of zeroes at the end of (127)! is

Q: The number of zeroes at the end of (127)! is

(A) 31

(B) 30

(C) 0

(D) 10

Sol. Number of zeroes

$\large = [\frac{127}{5}] + [\frac{127}{25}] + [\frac{127}{125}]$

= 25 + 5 + 1

= 31

Hence (A) is the correct answer.