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Solved Objective Question on Permutations and Combinations Set 1

Posted on - 15-05-2017

JEE Math PC

IIT JEE

Question.1

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

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 – 2nCn

(B) 2n

(C) 2n – 1

(D) 3n

Solution

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 Ç Q. Therefore, the number of ways in which none of a1, a2, . . .an 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

(C) 1

(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

(C) 30 24

(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 M1 and 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

(C) 6

(D) 18

Solution

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.

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

(C) 35

(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) 2n + 1

(C) 2n –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!

(C) 14C2

(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

(C) 41000

(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

(C) 1958

(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

 
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