Bunuel wrote:
There are x children and y chairs 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)?
(1) x + y = 12
(2) There are more chairs than children.
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:There are x children and y chairs.
x and y are prime numbers.
Statement 1: x + y = 12
Since x and y are prime numbers, a quick run on 2, 3, 5 shows that there are two possible cases:
Case 1: x=5 and y=7
There are 5 children and 7 chairs.
Case 2: x=7 and y=5
There are 7 children and 5 chairs
At first glance, they might look like two different cases and you might feel that statement 1 is not sufficient alone. But note that the question doesn’t ask you for number of children or number of chairs. It asks you about the number of arrangements.
Case 1: x=5 and y=7
If there are 5 children and 7 chairs, we select 5 chairs out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.
Total number of arrangements would be 7C5 * 5!
Case 2: x = 7 and y = 5
If there are 7 children and 5 chairs, we select 5 children out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.
Total number of arrangements would be 7C5 * 5!
Note that in both cases the number of arrangements is 7C5*5!. Combinatorics does not distinguish between people and things. 7 children on 5 seats is the same as 5 children on 7 seats because in each case you have to select 5 out of 7 (either seats or children) and then arrange 5 children in 5! ways.
So actually this statement alone is sufficient! Most people would not have seen that coming!
Statement 2: There are more chairs than people.
We don’t know how many children or chairs there are. This statement alone is not sufficient.
Answer: A.
We were tempted to answer the question as (C) but it was way too easy. Statement 1 gave 2 cases and statement 2 narrowed it down to 1. Be aware that if it looks too easy, you are probably missing something!