Lucky2783 wrote:

Question 2: There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

Statement 1: x + y = 12case 1 : x=7, y=5 ----> \(^7C_5 * (5-1)!\)

case 2: x=5, y=7 ----> as there are more chairs than there are children we cannot use circular permutation . there will be two empty chairs.

Choose 2 chairs which are empty out of 7 , #ways \(^7C_2\) and now the 5 children can be arranged in 5! ways .

note \(^7C_2\) = \(^7C_5\)

so total number of ways are \(^7C_5\) * \(5!\)

Not Sufficient.

Statement 2: There are more chairs than children

Insufficient.

together A and B.

we know this is case:2 of statement A .should be sufficient.

answer C

VeritasPrepKarishma please confirm whether this is correct .

Your answer is correct but your case 2 has a problem. If you have 7 distinct objects, then you can choose 5 in 7C5 ways. But here, the chairs arranged in a circle are identical initially. Say you choose two chairs right next to each other, how are they any different from any other two chairs right next to each other?

When you have 5 children and 7 chairs in a circle, you place the first child on any chair in 1 way.

Then you have 4 children and 6 distinct chairs. You can place the 4 children in 6*5*4*3 ways.

Total number of ways of placing 5 children in 7 chairs in a circle = 6*5*4*3

This is not the same as case 1 (7C5*4!).

So you need to know whether the number of chairs is more or the number of children is more. Hence you need both statement to answer the question.

Answer (C)

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