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Consider the expression [#permalink]
 07 Sep 2010, 06:24
Question Stats:
37% (02:04) correct
62% (02:13) wrong based on 0 sessions
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
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Last edited by shrouded1 on 07 Sep 2010, 06:37, edited 1 time in total.
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Re: Fun with factorials [#permalink]
07 Sep 2010, 06:33
Please paste the picture or use the math tags.... its difficult to understand the expression.
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Re: Fun with factorials [#permalink]
07 Sep 2010, 06:52
I checked it out with a few numbers and realized B is the answer. but am wondering how to solve this mathematically.
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Re: Fun with factorials [#permalink]
07 Sep 2010, 17:49
hemanthp wrote: I checked it out with a few numbers and realized B is the answer. but am wondering how to solve this mathematically. Please Explain!!!
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Re: Fun with factorials [#permalink]
07 Sep 2010, 22:04
I dont think this can be solved by me in actual GMAT.
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Re: Fun with factorials [#permalink]
08 Sep 2010, 01:27
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.
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Re: Fun with factorials [#permalink]
08 Sep 2010, 13:20
Can some one please provide a detailed explanation?
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Re: Fun with factorials [#permalink]
20 Oct 2010, 07:00
I give up on this question.
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Re: Fun with factorials [#permalink]
23 Oct 2010, 03:41
This one was so hard. Couldn't solve 
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Re: Fun with factorials [#permalink]
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. Posted from my mobile device 
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Re: Fun with factorials [#permalink]
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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Re: Fun with factorials [#permalink]
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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Re: Fun with factorials [#permalink]
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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