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There are x people and y chairs in a room where x and

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There are x people and y chairs in a room where x and

by M7MBA » Sat May 05, 2018 11:53 pm

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There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?

(1) x + y = 12

(2) There are more chairs than people.

The OA is the option A.

How can I show that the statement (1) is sufficient? Could someone clarify this to me? Please.
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by Vincen » Sun May 06, 2018 9:31 am

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Hello M7MBA.

Let's take a look at this question.

(1) x + y = 12

Since "x" and "y" are prime, then we have two options:

(i) x=5 and y=7.

First, we have to select 5 chairs from 7, this can be done with a combination: $$7C5=\frac{7!}{2!5!}=\frac{7\cdot6\cdot5!}{2\cdot5!}=21.$$ Now, we have to organize the 5 people in the 5 selected chairs, this can be done in 5! different ways. Therefore, the answer to the original question is: $$7C5\cdot5!=21\cdot5!=2520.$$

(ii) x=7 and y=5.

First, we have to select 5 people from 7, this can be done with a combination: $$7C5=\frac{7!}{2!5!}=\frac{7\cdot6\cdot5!}{2\cdot5!}=21.$$ Now, we have to organize the 5 selected people in the 5 chairs, this can be done in 5! different ways. Therefore, the answer to the original question is: $$7C5\cdot5!=21\cdot5!=2520.$$

Since both answers are the same, we can conclude that this statement is SUFFICIENT.

(2) There are more chairs than people.

This doesn't give enough information. Therefore, this statement is NOT SUFFICIENT.

Finally, the correct answer is the option A .

I hope it helps.
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