The problem in question is College Scholarship:

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.

The answer says that it should be both together are sufficient but neither alone is sufficient. I contend however that both alone are sufficient.

From my calculations if just given (1) then to calculate the total different number of ways to dole out the scholarships you would use (10 choose 6)*(3^6) which is 153090 ways. If just given (2) one would calculate by using that the committee “will grant” scholarships you would use (10 choose 3)*(3!) + (10 choose 6)*(6!) + (10 choose 9)*(9!) which is 3780720. Please review my work to double check it.

ah, ok, i see what you're doing here. unfortunately, your reasoning - while totally rational - is at odds with the accepted meaning of the terms in the problem statement. let me explain.

the problem statement says that the committee will give out "a certain number of $1k, $5k, and $10k scholarships".
this means that the numbers of each type of scholarship are FIXED (although they are unknown at the outset of the problem).

your answer to problem (1) takes into account EVERY different way in which you could have 6 of those scholarships, total.
while i see what you are doing, this is not the accepted meaning of the term "a certain number of".

same problem with your treatment of (2). in that problem, you are taking the number of ways of giving out 1 of each scholarship, and then adding that to the number of ways of giving out 2 of each, and then adding that to the number of ways of giving out 3 of each.
that's not what "a certain number of" means.

finally, 153,090 does not equal 3,780,720. since the two statements in a data sufficiency problem can NEVER contradict each other, this observation proves that you have interpreted the statements incorrectly.
i.e., if the two statements are each sufficient to determine a specific # for some quantity, then they MUST determine the same number. always.

Pls explain how to solve this problem....in case it was given in the Problem Solving category instead of the Data Sufficiency category.

well, you can't just take a data sufficiency problem and decide that you're going to call it a problem solving problem; you have to tweak it a bit. in particular, you have to take the statements that are "sufficient" (obviously, since otherwise you have a problem that can't be solved).

so, since the answer to the DS problem is (c), we have to assume that we actually have both of the statements together if we're going to write a "PS version" of this problem.

so the new problem is
if two of each scholarship type are granted, then in how many ways can the scholarships be disbursed?
(this is what you get if (a) there are 6 scholarships and (b) the 6 are divided equally among the categories)

one way to do this is with the anagram grid.
let's use "B" for the two big scholarships, "M" for the middling ones, and "S" for the small ones. then, there are four students who don't get anything - these will be "N"s.
your anagram is BBMMSSNNNN
so that's (10!) / (2!2!2!4!)
that number is your answer.

you can also use the slot method.
six slots.
the first two are the ways to disburse the big scholarships. there are 10 and 9 ways to choose the recipients (since the same student can't get both of them).
then, the next two slots for the middling scholarships - 8 and 7 respectively.
finally, the two slots for the small scholarships - 6 and 5 respectively.
since "order doesn't matter" between the two of each type of scholarship, you divide by three "2!"s.
therefore, your answer is (10x9x8x7x6x5) / (2!2!2!), which is the same number as above (in a slightly different form).

Ahhh....so that's how one can solve it...thanks...I tend to get a lil confused whr perm n combination problems r involved. :-)

(P.S - I assumed that it was implied that one has to take both statements of DS into account as d answer was C...will be more specific next time...thanks again)

Glad it helped!

_________________
Ben Ku
Instructor
ManhattanGMAT

Can you please explain why answer is 10!/2!2!2!4! and not 10!/2!2!2!4!6! why the order matters for the team

hi -



please take the extra fraction of a second to type out words such as "the". thank you in advance.

there may be posters reading this thread who think you're actually trying to say that answers (d) and (c) are the same.

hi -



the order doesn't matter within any of the individual groups.
hence, the division by the 2!'s (for each pair of recipients, who can be shuffled), and by 4! (for the set of four non-recipients, who can also be shuffled).

there is no group of 6 all of whom may be shuffled, so you can't divide by 6.

if you were just giving out 6 undifferentiated scholarships, then you would scratch the 2!2!2! and replace it with 6!, yielding 10!/(4!6!). perhaps this is what you are thinking about.

--

also, 10!/2!2!2!4!6! will never be a correct expression in any combinatorics problem, since 2 + 2 + 2 + 4 + 6 is greater than 10. you will never divide by factorials whose numbers add to more than the total set that you have to begin with.

(...and 10!/2!2!2!4!6! is not even a whole number)

RonPurewal,

2 questions please:

1) I think I can solve this question correct using the anagram method. Can you please make sure that my logic is correct?
Simply put, you need to figure out the total number of ways to distribute the scholarships (order does not matter between the 2 students who receive the same scholarship, e.g., $10K to one student and $10K to the other - not distinguishable from one another). Hence, you divide the total factorial (10!) by two factorials: one for the chosen group and one for those not chosen. Since we have 3 groups of 2 that are the "chosen ones," we place 2!2!2! multiplied by 4! (4! being the not chosen ones) in the denominator. Is my thought process sound?

2) Perhaps the more important question I have for you is how to solve this problem using the combinations formula: n!/[(n-r)!*r!] where r items are chosen from a pool of n items. For the vast majority of combinatorics problems, I use the combinations/permutations formula (as opposed to the anagram method) by first differentiating whether order matters vs. order doesn't matter. I personally find this approach much more intuitive. Do you advise against this? In other words, are there certain combinatorics problems that I should use the comb./perm. formulas vs. the anagram method?

Thank you very much,
Julgi.

1) Your use of the anagram method is correct.

2) for simple problems, the anagram method correctly applied will yield the same expression as will a correct translation directly into the combinatoric --n!/(n-r)!r!--or permutation--n!/(n-r)!. Think of the anagram method as a tool to translate the problem. If, as you say, you can generally translate as accurately and more comfortably without using the anagram method, that's great. This question does point out, though, that even for you the anagram method has a place.

There's more than one way to translate this without using something like the anagram method, but most such ways will look something like this (I'm using both statements together here):

First, determine the number of ways to award the two $10,000 scholarships, 10!/8!2!

Second, determine the number of ways to award the two $5,000 scholarships among the remaining eight students, 8!/6!2!

Third, determine the number of ways to award the two $1,000 scholarships among the remaining six students, 6!/4!2!

Finally, relying on the fundamental counting principle, take the product of those three values, (10!/8!2!)(8!/6!2!)(6!/4!2!)=10!/2!2!2!4!

Just to be clear, the message here is that you don't want to do this. USE THE ANAGRAM METHOD for questions like this one.