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A college admissions committee will grant a certain number

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A college admissions committee will grant a certain number  [#permalink]

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New post 12 Apr 2010, 13:56
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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.

MY QUESTION IS:

IS IT NOT POSSIBLE TO FIND THE ANSWER WITHOUT EVEN TAKING ANY STATEMENT INTO CONSIDERATION..??
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New post 13 Apr 2010, 02:27
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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?

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 \(xxyyzz\).

Answer: C.

Hope it's clear.
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Re: College Adminssion  [#permalink]

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New post 12 Apr 2010, 18:26
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This question is confusingly worded. I suspect your source is not an official GMAT exam. However, I will make an attempt.

T= number of 10,000 scholarships
F= number of 5,000 scholarships
O= number of 1,000 scholarships

The question stem tells us there will be three types of scholarships awarded to 10 students; however, we don't know how many of each type of scholarship will be awarded, nor do we know how many students will get scholarships. However we know that between 3 and 10 scholarships will be awarded.

S1. We know six scholarships will be awarded, so T+F+O = 6. However it could be many different combinations:
4T+1F+1O = 6 and the combinations would be 10C4*6C1*5C1
3T+2F+1O = 6 and the combinations would be 10C3*7C2*5C1
Insuff

S2. Now we know that we have either 1T+1F+1O, or 2T+2F+2O, or 3T+3F+3O, but we don't know which.
Insuff

Together:
It must be 2T+2F+2O
So the combinations are 10C2*8C2*6C2
Answer C
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Re: College Adminssion  [#permalink]

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New post 13 Apr 2010, 04:38
Bunuel wrote:
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 \(xxyyzz\).

Answer: C.

Hope it's clear.



Hello Brunel

If we just look at the given statement and let us say its a problem Solving (PS) question..!!!
Do we really need extra information to crack this question..

"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"
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New post 13 Apr 2010, 05:54
nverma wrote:
Hello Brunel

If we just look at the given statement and let us say its a problem Solving (PS) question..!!!
Do we really need extra information to crack this question..

"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"


OK, I see. "A certain number of $10,000, $5,000, and $1,000 scholarships" means that the numbers of each type of scholarship are FIXED. We just don't know this numbers.

If we were asked to calculate # of ways to distribute all possible numbers of scholarships (<=10), with all possible breakdowns among them (10-$10,000, 0-$5,000, 0-$1,000; 9-$10,000, 1-$5,000, 0-$1,000; ... 3-$10,000, 1-$5,000, 5-$1,000 ... huge # of combinations), then we could calculate it. But this is not what the question is asking.
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Re: College Adminssion  [#permalink]

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New post 13 Apr 2010, 16:11
Bunuel wrote:
nverma wrote:
Hello Brunel

If we just look at the given statement and let us say its a problem Solving (PS) question..!!!
Do we really need extra information to crack this question..

"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"


OK, I see. "A certain number of $10,000, $5,000, and $1,000 scholarships" means that the numbers of each type of scholarship are FIXED. We just don't know this numbers.

If we were asked to calculate # of ways to distribute all possible numbers of scholarships (<=10), with all possible breakdowns among them (10-$10,000, 0-$5,000, 0-$1,000; 9-$10,000, 1-$5,000, 0-$1,000; ... 3-$10,000, 1-$5,000, 5-$1,000 ... huge # of combinations), then we could calculate it. But this is not what the question is asking.



Thanks Brunel..

I feel if the Question given is DS.. we are supposed to use the options..!! :) :)
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Re: College Adminssion  [#permalink]

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New post 05 Aug 2010, 09:45
nverma wrote:
Bunuel wrote:
nverma wrote:
Hello Brunel

If we just look at the given statement and let us say its a problem Solving (PS) question..!!!
Do we really need extra information to crack this question..

"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"


OK, I see. "A certain number of $10,000, $5,000, and $1,000 scholarships" means that the numbers of each type of scholarship are FIXED. We just don't know this numbers.

If we were asked to calculate # of ways to distribute all possible numbers of scholarships (<=10), with all possible breakdowns among them (10-$10,000, 0-$5,000, 0-$1,000; 9-$10,000, 1-$5,000, 0-$1,000; ... 3-$10,000, 1-$5,000, 5-$1,000 ... huge # of combinations), then we could calculate it. But this is not what the question is asking.

Thanks Brunel..
I feel if the Question given is DS.. we are supposed to use the options..!! :) :)


I too was confused with this question. Initially I thought there are 10 students and three kinds of scholarships. So for each student there are 4 options - no scholarship or one of the three scholarship. This was giving me
4^10 possibiliites.
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Re: College Adminssion  [#permalink]

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New post 05 Oct 2010, 15:49
I'm still confused by this problem (1 out of 2 I got wrong on MGMAT CAT #6). I picked A, but I was going back and forth between A and C.

For A, let's say we have 6 scholarships, w/ 3 types. Couldn't we branch off the possibilities as:
1 $10k, 4 $5k, and 1 $1k.
2 $10k, 3 $5k, and 1 $1k.
etc.

Calculate all those possibilities, add them up, and we have an answer. There'll be a very large number of possibilities to hand out 6 scholarships of varying types, but in my mind, that figure DOES answer the question.

Why am I incorrect in assuming this?
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Re: College Adminssion  [#permalink]

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New post 05 Oct 2010, 16:11
NewSc2 wrote:
I'm still confused by this problem (1 out of 2 I got wrong on MGMAT CAT #6). I picked A, but I was going back and forth between A and C.

For A, let's say we have 6 scholarships, w/ 3 types. Couldn't we branch off the possibilities as:
1 $10k, 4 $5k, and 1 $1k.
2 $10k, 3 $5k, and 1 $1k.
etc.

Calculate all those possibilities, add them up, and we have an answer. There'll be a very large number of possibilities to hand out 6 scholarships of varying types, but in my mind, that figure DOES answer the question.

Why am I incorrect in assuming this?


It is a poorly worded question, what you are saying is not incorrect. But what the question means to get at is that the number of ways is governed by how many types of each scholarship there are.

Going by the same logic as yours, we don't even need statement A. We can just enumerate all the ways to give 0,1,2,3,4,... scholarships

All in all I think it is the question which is ambiguous.
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Re: College Adminssion  [#permalink]

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New post 05 Oct 2010, 16:12
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NewSc2 wrote:
I'm still confused by this problem (1 out of 2 I got wrong on MGMAT CAT #6). I picked A, but I was going back and forth between A and C.

For A, let's say we have 6 scholarships, w/ 3 types. Couldn't we branch off the possibilities as:
1 $10k, 4 $5k, and 1 $1k.
2 $10k, 3 $5k, and 1 $1k.
etc.

Calculate all those possibilities, add them up, and we have an answer. There'll be a very large number of possibilities to hand out 6 scholarships of varying types, but in my mind, that figure DOES answer the question.

Why am I incorrect in assuming this?


Question is about distributing the scholarships among 10 students.

Now, suppose there are 3 - 10K scholarships (A), 2 - 5K scholarships (B), and 1 - 1K scholarships (C), total of 6 scholarships. Then:

Students:
1-2-3-4-5-6-7-8-9-10
A-A-A-B-B-C-N-N-N-N
N-A-A-B-B-C-A-N-N-N
N-N-A-B-B-C-A-A-N-N
...
# of ways to distribute 6 scholarships with this particular breakdown among them (3-2-1) would be \(\frac{10!}{3!2!4!}\) (N - no scholarship);

But if: there are 1 - 10K scholarships (A), 1 - 5K scholarships (B), and 4 - 1K scholarships (C), again total of 6 scholarships. Then:

Students:
1-2-3-4-5-6-7-8-9-10
A-B-C-C-C-C-N-N-N-N
N-A-B-C-C-C-C-N-N-N
N-N-A-B-C-C-C-C-N-N
...
# of ways to distribute 6 scholarships with this particular breakdown among them (1-1-4) would be \(\frac{10!}{4!4!}\);

So for different breakdown of 6 scholarships (3-2-1 and 1-1-4) we've got different # of distributions (also different pattern), so knowing # of scholarships is not sufficient to answer the question, we should also know the breakdown among them.

Hope it's clear.
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New post 02 Apr 2011, 17:24
Bunuel wrote:
(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 \(xxyyzz\).

Answer: C.

Hope it's clear.


Hi Bunuel -

The equation that I came up with using the information from Statements 1 and 2 is \(\frac{10!}{2!2!2!4!}\) which is different from what you have \(\frac{6!}{2!2!2!}\). Can you explain please? Thank you so much in advance!
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New post 03 Apr 2011, 00:39
MarianneG wrote:
Bunuel wrote:
(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 \(xxyyzz\).

Answer: C.

Hope it's clear.


Hi Bunuel -

The equation that I came up with using the information from Statements 1 and 2 is \(\frac{10!}{2!2!2!4!}\) which is different from what you have \(\frac{6!}{2!2!2!}\). Can you explain please? Thank you so much in advance!


MarianneG wrote:
Bunuel wrote:
(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 \(xxyyzz\).

Answer: C.

Hope it's clear.


Hi Bunuel -

The equation that I came up with using the information from Statements 1 and 2 is \(\frac{10!}{2!2!2!4!}\) which is different from what you have \(\frac{6!}{2!2!2!}\). Can you explain please? Thank you so much in advance!


Actually, your answer matches with Bunuel's. You perhaps overlooked the multiplication with the given combination.

Bunuel's answer is: \(C^{10}_{6}*\frac{6!}{2!2!2!}=\frac{10!}{6!4!}*\frac{6!}{2!2!2!}=\frac{10!}{2!2!2!4!}\)
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Each year, a college admissions committee grants a certain  [#permalink]

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New post 11 Jul 2011, 00:46
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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?

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

(2) An equal number of scholarships will be granted at each scholarship level.
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New post 11 Jul 2011, 03:22
mojorising800 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?

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

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



st1: this can 10C4*6C1*5C1 or 10C2*8C2*6C2.......... insufficient

st2: this can be 10C2*8C2*6C2 or 10C3*7C3*4C3.........insufficient


1&2 : its 10C2*8C2*6C2.
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Re: A college admissions committee will grant a certain number  [#permalink]

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New post 28 Sep 2015, 04:03
Bunuel 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?

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 \(xxyyzz\).

Answer: C.

Hope it's clear.


I am getting confused with the \(C^6_{10}\). Shouldn`t the 10 be on the top and 6 on the bottom?
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Re: A college admissions committee will grant a certain number  [#permalink]

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New post 28 Sep 2015, 04:12
chingunee wrote:
Bunuel 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?

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 \(xxyyzz\).

Answer: C.

Hope it's clear.


I am getting confused with the \(C^6_{10}\). Shouldn`t the 10 be on the top and 6 on the bottom?


it is already defined at the top that \(C^n_{10}\) = choosing n students out of 10.

Also as you can not have negative factorials, there is only 1 interpretation for \(C^6_{10}\) =\(\frac{10!}{4!*6!}\)
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Re: A college admissions committee will grant a certain number  [#permalink]

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New post 31 Mar 2016, 05:54
Please let me know where I am going wrong.

1. I selected 6 students by 10C6 and then all these students can be given 3 types of scholarships.
So answer comes out to be 10C6*3^6.

2.FOR STATEMENT 2,
as equal number of scholarship will be granted at each level,
It can either be 3,3,3 or 2,2,2 or 1,1,1.
So answer comes out to be:
10c3*7c3*4c3 + 10c2*8c2*6c2 + 10c1*8c1*6c1.


Please help..!!!

Posted from my mobile device
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New post 04 Jun 2016, 04:36
why A is wrong here?

Even if we do not know the break up of each scholarship ,total number os scholarships should give us a definite answer on number of ways 6 scholarships can be given ?
# ways = 10 * 9 * 8 *7 &6 * 5 * 4

where am i going here ? please explain.

Thanks for your help
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New post 21 Jun 2016, 07:58
I agree that the answer is C. However, i dont arrive to the same answer as Bunnuel´s.

Ans: (10C8x8C6x6C4)/3!

The only difference with Bunnuel´s answer is that he did not divide the whole equation by 3!. However, i believe this is necessary because we are making 3 groups of 2, and it is the same to have peter and john with the scholarship of 10.000 and to have John and Peter with that scholarship

Thanks in advance to whoever helps me with this doubt!
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Re: A college admissions committee will grant a certain number  [#permalink]

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New post 11 Jul 2018, 10:24
nverma 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: Total 6 scholarships granted, hence we need 6 students to receive these scholarships.

6 students can be chosen out of 10 students in 10C6 ways.

However we do not know which & how many of each scholarships are granted.

Statement 1 is Not Sufficient.



Statement 2: Equal number of scholarships from each level are granted.

We do not know the exact number of each scholarship from each level.

Statement 2 is Not Sufficient.


Combining we get, 6 scholarships given out & equal # of each of the 3 levels. Hence 2 scholarships from each level.

Now 6 students are selected out of 10 in 10C6 ways.

We can select 2 students out of the 6, to receive 2 scholarships in 6C2

Similarly 2 from remaining 4 students in 4C2 ways

& lastly we give out the final 2 scholarships to the 2 remaining students in 2C2 ways.

Total # of ways to give out the scholarships = 10C6 * 6C2 * 4C2 * 2C2


Combining is Sufficient.



Answer C.



Thanks,
GyM
Re: A college admissions committee will grant a certain number &nbs [#permalink] 11 Jul 2018, 10:24
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