Data Sufficiency

niketdoshi123 wrote:Each year, a college admissions committee grants a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. The number of scholarships granted at each level does not vary from year to year, and no student can receive more than one scholarship. This year, how many different ways can the committee distribute the scholarships among the pool of 10 applicants?

a)In total, six scholarships will be granted.
b)An equal number of scholarships will be granted at each scholarship level.

Let us assume that number of $10,000 scholarships = A,
Number of $5,000 scholarships = B,
Number of $1,000 scholarships = C, and
Total number of scholarships granted = T
Then T = A + B + C
Number of different ways the committee can distribute scholarships among the pool of 10 applicants = 10CT * (A + B + C)!/(A! * B! * C!) = 10CT * T!/(A! * B! * C!)

Now we need to know the values of A, B, C, and T to answer the question.

(1) In total, six scholarships will be granted implies T = 6, but we don't know A, B, and C; NOT sufficient.

(2) An equal number of scholarships will be granted at each scholarship level. implies A = B = C, but again we don't know A, B, and C; NOT sufficient.

Combining (1) and (2), T = 6, A = B = C
T = A + B + C implies 6 = A + B + C, which implies that A = B = C = 2

So, required number of ways = 10CT * T!/(A! * B! * C!) = 10C6 * 6!/(2! * 2! * 2!); SUFFICIENT.

The correct answer is C.