This question is simply asking us to come up with the number of permutations that can be formed when x people are seated in y chairs. It would seem that all we require is the values of x and y. Let's keep in mind that the question stem adds that x and y must be prime integers.

(1) SUFFICIENT: If x and y are prime numbers and add up to 12, x and y must be either 7 and 5 or 5 and 7. Would the number of permutations be the same for both sets of values?
Let's start with x = 7, y = 5. The number of ways to seat 7 people in 5 positions (chairs) is 7!/2!. We divide by 2! because 2 of the people are not selected in each seating arrangement and the order among those two people is therefore not significant. An anagram grid for this permutation would look like this:
A B C D E F G
1 2 3 4 5 N N
But what if x = 5 and y = 7? How many ways are there to position five people in 7 chairs? It turns out the number of permutations is the same. One way to think of this is to consider that in addition to the five people (A,B,C,D,E), you are seating two ghosts (X,X). The number of ways to seat A,B,C,D,E,X,X would be 7!/2!. We divide by 2! to eliminate order from the identical X's.

(2) INSUFFICIENT: This statement does not tell us anything about the values of x and y, other than y > x. The temptation in this problem is to think that you need statement 2 in conjunction with statement 1 to distinguish between the x = 5, y= 7 and the x = 7, y = 5 scenarios.

ronnie1985 wrote:48. 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.

OA [spoiler](A)[/spoiler]

Statement 1: x+y = 12.
Since x and y must be prime numbers, either x=5 and y=7 or x=7 and y=5.

Case 1: 5 people, 7 chairs
There are FEWER PEOPLE than chairs.
The result is that, while not every chair will hold a person, EVERY PERSON MUST CHOOSE A CHAIR.
Thus, we count the number of options for each PERSON.
Number of chair options for the 1st person = 7.
Number of remaining options for the 2nd person = 6.
Number of remaining options for the 3rd person = 5.
Number of remaining options for the 4th person = 4.
Number of remaining options for the 5th person = 3.
To combine these options, we multiply:
7*6*5*4*3.

Case 2: 7 people, 5 chairs
There are FEWER CHAIRS than people.
The result is that, while not every person will choose a chair, EVERY CHAIR MUST HOLD A PERSON.
Thus, we count the number of options for each CHAIR.
Number of people who could sit in the 1st chair = 7.
Number of remaining people who could sit in the 2nd chair = 6.
Number of remaining people who could sit in the 3rd chair = 5.
Number of remaining people who could sit in the 4th chair = 4.
Number of remaining people who could sit in the 5th chair = 3.
To combine these options, we multiply:
7*6*5*4*3.

Since in case the number of ways to seat the people is the same, SUFFICIENT.

Statement 2: There are more chairs than people
No way to determine how the people could be seated.
INSUFFICIENT.

The correct answer is A.