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Scholarships

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by Uri » Fri May 01, 2009 12:44 pm
A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants?

(1) In total, six scholarships will be granted.

(2) An equal number of scholarships will be granted at each scholarship level.

Please explain your logic.

OA: [spoiler](C)[/spoiler]
Source: MGMAT CAT
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Source: — Data Sufficiency |

by bluementor » Thu May 07, 2009 1:19 am
Let's say:

x = number of $10,000 scholarships
y = number of $5,000 scholarships
z = number of $1,000 scholarships

no. of applicants = 10.

Statement 1: In total, six scholarships will be granted.

we have 10 applicants and 6 scholarships, so we can choose 6 people from the 10 applicants in 10C6 ways.

if all 6 scholarships are different, then we will have 6! ways to distribute within a certain combination of 6 recipients. however, in this case, we only have 3 different scholarships and we are unsure of x, y and z.

if x=1=y=1 and z=4, a given set of 6 recipients can receive scholarship in 6!/4 ways.
if x=y=z=2, a given set of 6 recipients can receive scholarship in 6!/(2x2x2) ways.

both these cases will yield different answers. hence insufficient.

Statement 2: An equal number of scholarships will be granted at each scholarship level.

this means x=y=z. however don't know the total number of scholarships awarded. insufficient.

both statements together:

Since we have a total of 6 scholarships, and the number of scholarships at each level is equal, we have x=y=z =2.

so for a given set of 6 recipients, we can distribute scholarships among them in 6!/(2x2x2) ways.

and we can choose 6 recipients from 10 applicants in 10C6 ways.

So the number of ways we can distribute scholarships among the pool of 10 applicants is (10C6) x (6!/8) ways. Sufficient.

Choose C.

-BM-
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