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A college admissions committee will grant a certain number [#permalink]
15 Aug 2008, 02:03

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

69% (01:39) correct
31% (01:31) wrong based on 108 sessions

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

Re: Probability Tough One [#permalink]
15 Aug 2008, 02:49

GMBA85 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

statement 1 : we have total number of scholarships, but we dont have breakup of each type of scholarships. not suff statement 2 : we dont have total number of scholarships, not suff

combine : we have 2 scholarships of each type... Suff

Re: Probability Tough One [#permalink]
15 Aug 2008, 03:40

durgesh79 wrote:

GMBA85 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

statement 1 : we have total number of scholarships, but we dont have breakup of each type of scholarships. not suff statement 2 : we dont have total number of scholarships, not suff

combine : we have 2 scholarships of each type... Suff

C

Do we really need to know distribution of scholarship?

Re: Probability Tough One [#permalink]
15 Aug 2008, 06:20

nmohindru ,

We do need to know the number of each scholarship. If you have a variety, there are more options. For example: 3 @ $1k, 2 @ 5k and 1 @ 10k.

This could be \(C_10^3 + C_10_2 + C_10_1\) because you need to know how many ways each of the scholarships could be given. Compare this to:

All $5k scholarships would be \(C_10^6\). The different scholarships acts somewhat as a differentiator by order. I'm not sure what the final answer is for how many ways, but I'm pretty sure the answer is C to the DS question.

GMBA85 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

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: Probability Tough One [#permalink]
15 Aug 2008, 14:31

durgesh79 wrote:

GMBA85 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

statement 1 : we have total number of scholarships, but we dont have breakup of each type of scholarships. not suff statement 2 : we dont have total number of scholarships, not suff

combine : we have 2 scholarships of each type... Suff

C

Why is 1) insufficient ? With 1) only, there 10C6 * (3^6) ways of doling out the schols .. choose 6 students in 10C6 ways and each of the 6 can be given one of the 3 types of schol.

Re: Probability Tough One [#permalink]
23 Aug 2009, 23:24

I think this is a very important question concept wise. We always think that selecting r items from a collection of n items can be done in nCr ways. However, this is true only if r items are of the same type.

If r items are different within themselves, then we get more options since we have to multiply the possibilities of each of these subtypes:)

Re: Probability Tough One [#permalink]
24 Aug 2009, 23:19

flyingbunny wrote:

should be A

for 1), the different ways are 10P6, doesn't matter what scholarships are.

for 2), doesn't mean anything

Answer is A.

Hi Flyingbunny,

That was my take too before I got into the details. stmt1 is not sufficient because 10P6 will work only if the 6 items under consideration are of the SAME type. If we have different types (or varieties) then we will have more number of options for selection:)

eg. If there are 5 shirts of same color and you have to select 5 then you don't have many options but to go for the same color.

If there are 5 shirts of different colors and you have to select 5 then you have got more choices.

Try putting some numbers for different scholarship types and check the difference:)

Re: Probability Tough One [#permalink]
25 Aug 2009, 10:43

1

This post received KUDOS

Answer is C

\(X_1\times10000+x_2\times5000+x_3\times1000=Total Money Pool Value\)

1) Stmt 1: \(X_1+x_2+x_3=6\). Howerver, we do not know how many scholarships of each type were distributed. We can't assume any (\(x_1,X_2,X_3\)) such that \(X_1+x_2+x_3=6\), since the equation X_1\times10000+x_2\times5000+x_3\times1000=Total Money Pool Value might be violated. 2) Statement 2 : \(X_1=x_2=x_3=a\) not sufficient obviously...we do not know the total number of distributed scholarships and we do not know the total value of the money pool...hence we can't figure out the values...note that if we had the total value of money pool we could solve the problem.

Combined \(x_1+x_2+x_3=6\), \(x_1=x_2=x_3\)

From here x_1=x_2=x_3=2

Last edited by LenaA on 25 Aug 2009, 21:07, edited 1 time in total.

Although this is a DS question, I would like to find the value.

Combining statements 1 and 2, we have: 2 x 10k 2 x 5k 2 x 1k These have to be distributed among 10 students.

Distribute 2x10k among 10 students = ways of picking 2 students from 10 = 10C2 Distribute 2x5k among 8 students left = ways of picking 2 students from 8 = 8C2 Distribute 1x5k among 6 students left = ways of picking 2 students from 6 = 6C2 => 10C2*8C2*6C2 = 45*28*15 = 18,900

Have the feeling I'm doing something wrong. _________________

Re: Probability Tough One [#permalink]
16 Oct 2009, 13:57

i thought it was A, but my logic is just because there are 10 people and each person cant get the same scholorship doesnt fully define the problem. Because we know there are 3 different scholorships but dont know if its 6 of each or 2 2 2 than we cant fully calculate the total prob, thus you need statement 2. and the answer is C.

Re: Probability Tough One [#permalink]
16 Oct 2009, 18:51

Economist wrote:

flyingbunny wrote:

should be A

for 1), the different ways are 10P6, doesn't matter what scholarships are.

for 2), doesn't mean anything

Answer is A.

Hi Flyingbunny,

That was my take too before I got into the details. stmt1 is not sufficient because 10P6 will work only if the 6 items under consideration are of the SAME type. If we have different types (or varieties) then we will have more number of options for selection:)

eg. If there are 5 shirts of same color and you have to select 5 then you don't have many options but to go for the same color.

If there are 5 shirts of different colors and you have to select 5 then you have got more choices.

Try putting some numbers for different scholarship types and check the difference:)

I fell into the trap of answering A too, good explanation Economist, Thanks _________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Re: Probability Tough One [#permalink]
28 Oct 2010, 01:27

powerka, Even I got the answer C and tried to find the number of ways using the following way: [1] choose any 6 students= 10C6 [2] Now, these 6 prizes(2+2+2) can be distributed to 6 people in 6!/ (2!*2!*2!) (divinding by 2! since 2 prizes of each of 3 prizes are same)

Thus number of ways: [1] * [2] = 18,900

However, I am not sure if the number of ways is correct.

Please, Quant moderator throw more light on the answer. Have my GMAT in a few days...

Re: Probability Tough One [#permalink]
28 Oct 2010, 01:34

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

ptm30 wrote:

powerka, Even I got the answer C and tried to find the number of ways using the following way: [1] choose any 6 students= 10C6 [2] Now, these 6 prizes(2+2+2) can be distributed to 6 people in 6!/ (2!*2!*2!) (divinding by 2! since 2 prizes of each of 3 prizes are same)

Thus number of ways: [1] * [2] = 18,900

However, I am not sure if the number of ways is correct.

Please, Quant moderator throw more light on the answer. Have my GMAT in a few days...

x= number of 10,000 scholarships y= number of 5,000 scholarships z= number of 1,000 scholarships n= total number of scholarships granted = x+y+z = number of students who will receive scholarship (as no student can receive more than one scholarship)

# different ways the committee can dole out the scholarships among the pool of 10 applicants is \(C^n_{10}*\frac{(x+y+z)!}{x!y!z!}=C^n_{10}*\frac{n!}{x!y!z!}\).

\(C^n_{10}\) - choosing \(n\) students who will be granted the scholarship; \(\frac{n!}{x!y!z!}\) - # of ways we can distribute \(x\), \(y\) and \(z\) scholarships among \(n\) students.

So we need to know the values of \(n\), \(x\), \(y\) and \(z\) to answer the question.

(1) \(n=6\). Don't know \(x\), \(y\) and \(z\). Not sufficient. (2) \(x=y=z\). Don't have the exact values of \(x\), \(y\) and \(z\) . Not sufficient.

(1)+(2) \(n=6\) and \(x=y=z\). As \(n=6=x+y+z\) --> \(x=y=z=2\) --> # of ways \(C^n_{10}*\frac{n!}{x!y!z!}=C^6_{10}*\frac{6!}{2!2!2!}\). Sufficient.

And again: \(C^6_{10}\) - choosing \(6\) students who will be granted the scholarship; \(\frac{6!}{2!2!2!}\) - distributing \(xxyyzz\) (two of each) among 6 students, which is the # of permuatrions of 6 letters \(xxyyzz\).

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