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by niketdoshi123 » Thu Jun 28, 2012 11:29 pm
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.
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Source: — Data Sufficiency |

by Anurag@Gurome » Thu Jun 28, 2012 11:39 pm
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.
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by sandeep_thaparianz » Sat Jun 30, 2012 5:57 am
Beautifully explained above..C
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by bubbliiiiiiii » Mon Jul 02, 2012 4:08 am
Could you please explain this?
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!)
Regards,

Pranay
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by GMATGuruNY » Mon Jul 02, 2012 8:37 am
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.
This is a PIECES-OF-THE-PUZZLE question.

Statement 1: In total, six scholarships will be granted.
A piece is missing: Are the scholarships all of the SAME type or of DIFFERENT types?
INSUFFICIENT.

Statement 2: An equal number of scholarships will be granted at each scholarship level.
A piece is missing: How MANY of each type are being granted?
INSUFFICIENT.

Statements 1 and 2 together:
Now we have ALL OF THE PIECES of the puzzle: two of each type of scholarship are being granted.
Thus, the number of possible distributions can be determined.
SUFFICIENT.

The correct answer is C.
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