Integers x and y are both positive, and x > y. How many different com

KAPLAN OFFICIAL SOLUTION:

The first step in this problem is to determine what we are really being asked. If we want to select committees of y people from a group of x people, we should use the combinations formula, which is n!/[k!/(n-k)!]. Remember, in this formula n is the number with which we start and k is the number we want in each group. Thus, we can reword the question as what does x!/[y!(x-y)!] equal?

Statement 1 tells us how many committees of x-y people we can make from our initial group of x people. If we plug this information into the combinations formula, we get x!/[(x-y)!(x-(x-y))!] = 3,060. This can be simplified to x!/[(x-y)!(x-x+y))!] = 3,060, which in turn is simplified to x!/[(x-y)!y!] = 3,060. The simplified equation matches the expression in our question, and gives us a numerical solution for it. Therefore, statement 1 is sufficient.

Statement 2 tells us how many ways we can arrange a number of people. The formula for arrangements is simply n!. In this case we have x-y people, thus (x-y)! = 24. Therefore, x-y must equal 4. However, we have no way of calculating what x and y actually are. This means that we cannot calculate the number of combinations in our question. Statement 2 is insufficient. So our final answer choice for this Data Sufficiency question is answer choice (A) or (1), Statement 1 is sufficient on its own, but Statement 2 is not.

~Bret Ruber