A college admissions committee will grant a certain number

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?

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

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

(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..??

Bunuel · Accepted Answer

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.

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.

TheGmatTutor · Answer

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

nverma · Answer

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"

Bunuel · Answer

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.

NewSc2 · Answer

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?

shrouded1 · Answer

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.

Bunuel · Answer

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.

GyMrAT · Answer

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

Prasannathawait · Answer

C has to be the answer without solving.

By A we know the number of scholarships is 6 but we don't know the combinations for denominations. It could be 2,2,2 or 3,2,1, etc for $10000,$5000,and $1000.

By B we know an equal number of scholarships are there but it could be 2,2,2 or 2,2,2,2 etc.

Combining the two statements we know only one possibility is there 2,2,2.

KarishmaB · Answer

There are 3 types of scholarships - 10k, 5k and 1k
How many and of what type, we don't know.

We have 10 total students.

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

Ok, so we know that we need to select 6 of the 10 students. But how do we split the scholarships among them? (1-1-4) or (1-2-3) or (2-2-2) etc. We don't know. We need to know how many of each type there are.
Not sufficient.

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

But how many total will be granted? Is it (1-1-1) or (2-2-2) or (3-3-3)? We don't know.
Not sufficient

Both together, we know that we have 2 scholarships of each type. 
So for two 10k scholarships, select 2 students in 10C2 ways.
For two 5k scholarships, select 2 students in 8C2 ways.
Now for two 1k scholarships, select 2 students in 6C2 ways.

Total ways = 10C2 * 8C2 * 6C2

Answer (C)

imslogic · Answer

Hi Karishma:

In Bunuel's explaination, it says 6C10. Do we invert this to 10C6 or assume 6C6? I was confused and wanted to confirm.

Best regards

Posted from my mobile device

IanStewart · Answer

Karishma and Bunuel were taking different perspectives on how to count here, so their solutions are different. Bunuel was asking "how many ways can we first choose the 6 people who will get scholarships, and then once we've done that, in how many ways can we assign the two $10k scholarships, the two $5k scholarships, and the two $1k scholarships to those 6 people?" Karishma instead solved by asking "in how many ways can we first choose two people, from all ten of the available people, to give the $10k scholarship, then in how many ways from the 8 remaining can we choose 2 people to give the $5k scholarship, then from the remaining six in how many ways can we choose 2 to give the $1k scholarship?"

Both solutions are correct, but if you're comparing one solution to the other, they won't line up at all. I'd solve the way Karishma did -- that perspective is more intuitive to me, and the resulting calculation is simpler.

If you are reading Bunuel's solution, in real combinatorics, there are many different notations people use when they want to express "10 choose 6" (how many ways to pick 6 things from 10 when order does not matter). The most common by far is this:

\binom{10}{6}

but that's really inconvenient to type on the internet, so in GMAT forums, people most of the time write 10C6 instead. There are other notations sometimes used in advanced combinatorics - for example this

{}^{10}	ext{C}_6

is common. Before reading gmatclub, I had never seen anyone use the notation

ext{C}^{6}_{10}

which I've seen many times in Bunuel's posts, but so many different notations are used in different combinatorics texts that it might be a standard usage in some books that I haven't seen before. So when Bunuel wrote that (he didn't write "6C10"), he meant the same thing as "10C6".

NCC · Answer

Solution:
scholarships = students

1] 6 scholarships given, so 6 students given the scholarship, 4 not given
10C6 X (6!/x!y!z!)

10C6 --> selecting 6 from 10 students to give the money
6! --> we are looking at arranging the scholarships (1 for each student)
x,y,z = we know scholarships are of a certain kind - 3 types --> so we need to divide 6! by these

Since we dont know x,y,z --> not suff

2] 10Cn X (n!/x!y!z!)

not sufficient

Together] 10C6 X (6!/2!2!2!)

Ans - C

A_Nishith · Answer

scholarships = students

Statement 1: 
6 scholarships given, so 6 students given the scholarship, 4 not given
10C6 X (6!/x!y!z!)

10C6 --> selecting 6 from 10 students to give the money
6! --> we are looking at arranging the scholarships (1 for each student)
x,y,z = we know scholarships are of a certain kind - 3 types --> so we need to divide 6! by these

Since we dont know x,y,z --> not suff

Statement 2: 
 10Cn X (n!/x!y!z!)

not sufficient

Together: 
 10C6 X (6!/2!2!2!)

Ans - C

heyitsmepearl · Answer

Hi, why can't we use the partition method to find the number of ways in which T+F+O=6 can be distributed?

I did 10C6 to select 6 out of 10 students= 210 ways

And then used partition method-
(6+3-1) C (3-1) = 28 ways

So total ways to distribute scholarship = 210 x 28 = 5880 ways.

Where am I going wrong?

Bunuel · Answer

The partition formula applies when identical items are being divided into 3 groups without distinguishing which specific students get which type. But here the 6 scholarships go to  different students , so order (who gets which scholarship type) matters.

We must first pick 6 students (10C6) and then assign which of them get 10k, 5k, and 1k. Since x=y=z=2, the correct count is 10C6 * 6!/(2!2!2!), not 10C6 * 28.

luisdicampo · Answer

Deconstructing the Question

There are   10   applicants and three types of scholarships.

Each student can receive at most one scholarship.

We need to determine the number of ways to assign the scholarships.

Step-by-step

Statement (1): A total of   6   scholarships are granted.

This only gives

a+b+c=6

but does not fix how many of each type are given.

Statement (1):  Insufficient

Statement (2): An equal number of scholarships are granted at each level.

So

a=b=c

but the total is unknown.

Statement (2):  Insufficient

Using both statements:

a+b+c=6   and   a=b=c

Thus

a=b=c=2

This uniquely determines the number of ways to assign scholarships.

Answer: C

A college admissions committee will grant a certain number

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