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# Consider the expression

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Retired Moderator
Joined: 02 Sep 2010
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07 Sep 2010, 06:24
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Difficulty:

55% (hard)

Question Stats:

42% (00:46) correct 58% (01:42) wrong based on 12 sessions

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Consider the expression $$\frac{(A!)}{((B!)^x * (C!)^y * (D!)^z))}$$ where A,B,C,D,x,y,z are all positive integers >=1. Is this expression an integer ?

(1) B+C+D < A
(2) xB+yC+zD < A
[Reveal] Spoiler: OA

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Last edited by shrouded1 on 07 Sep 2010, 06:37, edited 1 time in total.

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08 Sep 2010, 01:27
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.
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23 Oct 2010, 03:55
It's a tricky one, but if you look at the solution above, it's almost like saying c(n,r) will always be an integer.

Just that in this case we are talking about a different kind of arrangement with a different formula.

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25 Oct 2010, 22:51
shrouded1 wrote:
It's a tricky one, but if you look at the solution above, it's almost like saying c(n,r) will always be an integer.

Just that in this case we are talking about a different kind of arrangement with a different formula.

Posted from my mobile device

indeed very tricky, i was able to rule out (1) and guessed B

Can you explain a bit more on what you mean by "it's almost like saying c(n,r) will always be an", i'm not seeing this thru a combination? Thanks.

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26 Oct 2010, 00:43
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.
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26 Oct 2010, 08:52
Nice, thanks for explaining, now i get the picture.

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Re: Fun with factorials   [#permalink] 26 Oct 2010, 08:52
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# Consider the expression

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