From CAT EXAM #2:

Medals are to be awarded to three teams in a 10-team competition. If one medal is gold, one medal is silver, and one medal is bronze, how many different ways are there to award the three medals to teams in the competition?

a) 10! / 7!
b) 10! / 3! 7!
c) 10! / 3!
d) 7! / 3!
e) 7! / 4! 3!

The correct answer is A, but I am a hard time understanding the explanation. I originally chose answer B. Can someone explain why the 3! in the denominator is not needed?

Thanks!

I don't know what the explanation says, but you are asking when to use permutations vs. combinations.

In this example, I think of each of the medals having the ability to choose a team.
Teams: A B C D E F G H I J
Medals: Gold, Silver, Bronze

Gold medal: 10 choices: A B C D E F G H I J. Let’s say A wins the Gold.
Silver medal: 9 choices: B C D E F G H I J. Let’s say B wins the silver.
Bronze medal: 8 choices: C D E F G H I J. Let’s say C wins the bronze.

Note: Order matters here. If Gold chooses A, it cannot choose B or C or D.

Thus, I picked certain teams to win: I had 10 choices at first, then 9, then 8. The total number of options was 10 * 9 * 8 = 720. Using the permutation formula this is written as 10!/(10-3)!

So when do we use the combinations formula? Assume that all the medals are gold. How many ways can I give 3 gold medals to 10 teams?

Well, in this case, the order we pick teams doesn’t matter. If I give a gold to Argentina, Brazil and Chile, it’s the same as giving it to Chile, Argentina, and then Brazil. Either way, they’re going to be equally happy.

So, if we have 3 gold medals to give away, there are 3! or 6 variations (ie ABC, ACB, BAC, BCA, CAB, CBA) for every choice we pick. If we want to figure out how many combinations we have, we just create all the permutations and divide by all the redundancies

This can be written using the combinations formula 10!/3!7!

hope that helps

Thanks for your reply! Very helpful!

hi.

i'm not sure of your level of exposure to our course materials, in particular the strategy guides, but i'm going to assume that you've read them (since you're a student).
in particular, i'll assume a passing familiarity with the "anagram method" and the "slot method", either of which may be used as a basis for solving this problem.

ANAGRAM METHOD:
you have one gold (G), one silver (S), one bronze (B), and seven non-recipients (N).
therefore, you're finding "anagrams" of the "word" GSBNNNNNNN
according to the anagram formula, this will be (10!) / (1!1!1!7!)
the 1!'s don't matter, so you're left with the formula in (a).

SLOT METHOD: (much easier)
you just have to distribute the three medals. since "order matters", totally (you aren't giving out more than one of anything), you just have to fill in the slots, and you're done.
gold medal: 10 ways to give
silver: 9 ways
bronze: 8 ways
therefore, 10 x 9 x 8.
that's the same as (10! / 7!).

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EDITOR'S NOTE:

i think it's much more likely that the answer choices would be listed as actual numbers (i.e., 720, rather than 10! / 7!) in the problem itself.
otherwise, the problem penalizes students who find the answer in the easier way (the slot method), and gmat problems are pretty much never set up to penalize students who find efficient solutions.

GMAT would have never been the same without you RON. Thanks a ton. Explanation helps me to differentiate between permutation and combination problem.
However, my learning from this problem is not to differentiate them, but just to follow the slot logic. :)

:)

_________________
Tim Sanders
Manhattan GMAT Instructor