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Re: Fun with factorials [#permalink]
08 Sep 2010, 00:27

The answer is B This is one of those questions me and a friend of mine made up to test each other (both of us have GMATs coming up). So I have a solution with me, but it is rather unconventional :

It is easy to prove that (1) alone cant be the answer since there is no constraint on x,y,z and one can make these big enough to exceed the numerator. Eg 7! / (2!^40 * 3!^20 * 1!^1)

To show that (2) alone is sufficient : Consider the question "How many permutations are possible of a set of A alphabets, of which x alphabets are each repeated B times, y alphabets each repeated C times and z alphabets each repeated D times within the set ?" The answer to this question is exactly the expression above, and we know that since it is the answer to a counting question, it must be an integer.

Re: Fun with factorials [#permalink]
23 Oct 2010, 02:41

This one was so hard. Couldn't solve

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Re: Fun with factorials [#permalink]
25 Oct 2010, 23:43

Imagine I ask you the question : What is the number of ways you can arrange A balls, each of different color in a row ? The answer would be A!

Now I modify that question : What is the number of ways you can arrange A balls, of which B are are blue, C are red, D are green and the rest are of different but unique colors ? The answer would now be \frac{A!}{B!C!D!}

Now I modify it further : What is the number of ways you can arrange A balls, of which there are x subsets each consisting B balls each such that each subset consists of balls of a different shade of blue, and all other balls not included in these subsets are of unique colors ? The answer would now be \frac{A!}{B!^x}. Notice that A has to be greater than or equal to xB

Finally I modify it a bit more : What is the number of ways you can arrange A balls, of which there are x subsets each consisting B balls each, y subsets of C balls each, z subsets of D balls each, such that each subset consists of a unique color of balls. And the rest of the set of balls are distinct from all other balls ? The answer would now be \frac{A!}{B!^xC!^yD!^z}. Notice that A has to be greater than or equal to xB+yC+zD

Essentially what I am getting at is that the expression shown above is the answer to a combinatorial problem. And since the answer to a combinatorial problem is a "number of ways", such an expression always has to be an integer.