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by SubratGmat2011 » Mon Sep 13, 2010 9:56 am
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.


Ans is C - Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient
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by sanju09 » Wed Sep 15, 2010 11:07 pm
SubratGmat2011 wrote: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.


Ans is C - Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient

There are p, q, and r number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships, respectively. Hence, there is a total of (p + q + r) ≤ 10 scholarships waiting to select and arrange any (p + q + r) applicants out of a pool of 10, which can be done in

10C(p + q + r) × (p + q + r)!/ (p! q! r!) number of ways.

Visibly, we must know p, q, r discretely before trying answer the stem here. Hence, the question in chief here is p, q, r = ?

(1) This only reads (p + q + r) = 6. Insufficient

(2) This reads p = q = r = 1, 2, or 3. Insufficient

Taken together, p = q = r = [spoiler]just 2. Sufficient


C[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by baiju09 » Wed Sep 15, 2010 11:34 pm
sanju09 wrote:
SubratGmat2011 wrote: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.


Ans is C - Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient

There are p, q, and r number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships, respectively. Hence, there is a total of (p + q + r) ≤ 10 scholarships waiting to select and arrange any (p + q + r) applicants out of a pool of 10, which can be done in

10C(p + q + r) × (p + q + r)!/ (p! q! r!) number of ways.

Visibly, we must know p, q, r discretely before trying answer the stem here. Hence, the question in chief here is p, q, r = ?

(1) This only reads (p + q + r) = 6. Insufficient

(2) This reads p = q = r = 1, 2, or 3. Insufficient

Taken together, p = q = r = [spoiler]just 2. Sufficient


C[/spoiler]
Catch is: translate the phrases to alphabets, but this one is brilliantly done, may I expect a better shot, experts?
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by Adi_Pat » Tue Sep 21, 2010 3:29 pm
SubratGmat2011 wrote: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.


Ans is C - Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient
Can someone plz clarify this...

we have ten students...and no student can get more than one scholarship...so for e.g....if we have two scholarships....

the committee could give the first one to either of the ten candidates...
and the second one to the remaining nine candidates...
Total no. of ways = 10* 9

So we basically need to know what the number of scholarships be handed out.

The first option gives us this information = 6 ...so total number of ways = 10*9*8*7*6*5...Hence this should be sufficient.

The second option says they are equal in number so we could have 3 or 6 or 9 scholarships...not Sufficient

What am I missing here ??
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by sanju09 » Thu Sep 23, 2010 10:25 pm
Adi_Pat wrote:
SubratGmat2011 wrote: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.


Ans is C - Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient
Can someone plz clarify this...

we have ten students...and no student can get more than one scholarship...so for e.g....if we have two scholarships....

the committee could give the first one to either of the ten candidates...
and the second one to the remaining nine candidates...
Total no. of ways = 10* 9

So we basically need to know what the number of scholarships be handed out.

The first option gives us this information = 6 ...so total number of ways = 10*9*8*7*6*5...Hence this should be sufficient.

The second option says they are equal in number so we could have 3 or 6 or 9 scholarships...not Sufficient

What am I missing here ??

To answer many different ways that the committee can dole out the scholarships among 6 out of a pool of 10, we first need to select those 6, this can be done in 10C6 = 210 ways. We would then arrange 6 scholarships of 3 different denominations among the selected 6, here comes a question mark! Had those scholarships been of 6 different denominations, this could have easily done in 6! = 720 ways, but the case is not thus. We can sense a repetition of elements out here, that's why (p + q + r)!/ (p! q! r!) kind of scene surfaces up in the minds, whose denominator (p! q! r!) openly demands p, q, r discretely known first.


Question to you now, can the statement 1 segregate it's 6 into 3 groups?
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

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