Solved Objective Question on Permutations and Combinations Set 1

Question.1

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

(A) 300

(B) 325

(D) 350

Solution

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 x 25 = 325

Question.2

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 ÇQ = f is.

(A) 2²ⁿ – ²ⁿC_n

(B) 2ⁿ

(D) 3ⁿ

Solution

Let A = { a₁, a₂, a₃, . . . , a_n} . For a_i ∈ A, we have the following choices:.

(i) a_i ∈ P and a_i∈Q (ii) a_i ∈ P and a_i∈Q

(iii) a_i ∈P and a_i∈Q (iv) a_i ∈ P and a_i∈Q

Out of these only (ii) , (iii) and (iv) imply a_i ∈ P Ç Q. Therefore, the number of ways in which none of a₁, a₂, . . .a_n belong to P Ç Q is 3ⁿ . .

Question.3

Let p be a prime number such that p ≥ 3. Let n = p! + 1. The number of primes in the list n+1, n+2, n+ 3, . . . . n + p –1 is .

(A) p –1

(B) 2

(D) none of these

Solution

For 1 £ k £ p –1, n+ k = p! +k +1, is clearly divisible by k +1.

Therefore, there is no prime number in the given list.

Question.4

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

(D) none of these.

Solution

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 M₁and W₁ or M₁ and W₂. Thus the number of ways of arranging the game is.

(⁹C₂) (⁷C₂)(2) = = 1512.

Question.5

Number of even divisions of 504 is

(A) 12

(B) 24

(D) 18

Solution

504 = 2³ ´ 3² ´ 7.. Any even divisor of 504 is of the form 2ⁱ ´ 3ⁱ ´ 7^k, where I £ i£ 3, 0 £ j £ 2, 0 £ k£ 1. Thus total number of even divisors is 3 ´3´ 2 = 18.

Question.6

Ten persons are arranged in a row. The number of ways of selecting four persons so that no two persons sitting next to each other are selected is.

(A) 34

(B) 36

(D) none of these

Solution

To each selection of 4 persons we associated a binary sequence of the form 1001001010 where 1(0) at ith place means the ith person is selected (not selected).

There exists one-to-one correspondence between the set of selections of 4 persons and set of binary sequence containing 6 zeros and 4 ones.

We are interested in the binary sequences in which 2 ones are not consecutive. We first arrange 6 zeros.

0 0 0 0 0 0

This can be done in just one way. Now, 4 ones can be arranged at any of the 4 places marked with a cross in the following arrangement.

´ 0 ´ 0 ´ 0 ´ 0 ´ 0 ´ 0.

We can arrange 4 1’s at 7 places in ⁷C₄ = 35 ways.

Question.7

A set contains (2n + 1) elements. The number of subset of the set which contain at most n elements is .

(A) 2ⁿ

(B) 2^{n + 1}

(D) 2²ⁿ

Solution

The number of subsets of the set which contain at most n elements is

(say)

we have

Question.8

No. of different garlands using 3 flowers of one kind and 12 flowers of second kind is .

(A) 19

(B) 11! ´ 2!

(D) none of these

Solution

Number of different garlands will be equal to number of different solutions of the equation a + b + c = 12 without taking order of a, b and c into consideration. .

Question.9

The number of ways in which letters of the word ARGUMENT can be arranged so that both the corners are filled by consonants only is

(A) 3! ´ 5!

(B) 14400

(D) None of these

Solution

Vowels of the word ARGUMENT divide the consonants in four parts, whose numbers are say y₁, y₂, y₃ and y₄ where y₁ + y₂ + y₃ + y₄ = 5,
y₁ ≥ 1, y₄ ≥1, y₂ ≥ 0, y₃ ≥0

Number of solutions of this equation is 20. Hence required number is 20 ´ 5!´ 3! = 864000 (as consonants can be arranged in 5! ways and vowels can be arranged in 3! Ways).

Question.10

The number of signals that can be generated by using 6 differently coloured flags, when any number of them may be hoisted at a time is

(A) 1956

(B) 1957

(D) 1959

Solution

When one flag is used, the number of signals that can be generated is ⁶P₁. When two flags are used, the number of signals that can be generated is ⁶P₂. when three flags are used, the number of signals that can be generated is ⁶P₃ and so on. Hence, the number of different signals that can be generated is.

⁶P₁ + ⁶P₂ +⁶P₃ +⁶P₄ +⁶P₅ +⁶P₆= 6 + 30 + 120 + 360 + 720 + 720 = 1956