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Larry, Michael, and Doug have five donuts to share. If any one of the [#permalink]
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07 Aug 2009, 05:57
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Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed? (A) 21 (B) 42 (C) 120 (D) 504 (E) 5040
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Re: Larry, Michael, and Doug have five donuts to share. If any one of the [#permalink]
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07 Aug 2009, 06:51
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OA AThis is little trivial rule in combinations The formula for distributing n identical things amongst r people such that any person might get any number of things ( including 0) is (n+r1)C(r1)in this problem n=5 & r=3 hence 7C2 = 21 ways **PS : if you want no of choices excluding 0 then => (n1)C(r1) = 4C2 = 6 ways
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Re: Larry, Michael, and Doug have five donuts to share. If any one of the [#permalink]
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07 Aug 2009, 06:59
question is not clear tho, if any one men can take from 0 to 5 donuts, is it allowed that together they only take less than 5? as in one guy take 1, the other take nothing. well that would make this a lil more complicated...



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Re: Larry, Michael, and Doug have five donuts to share. If any one of the [#permalink]
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22 Sep 2009, 15:07
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bhushan252 wrote: OA A
The formula for distributing n identical things amongst r people such that any person might get any number of things ( including 0) is (n+r1)C(r1) Good to know ... +1 for that Btw, this is a Manhattan Gmat question. Manhattan's explanation is 1 page long.
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Re: Larry, Michael, and Doug have five donuts to share. If any one of the [#permalink]
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22 Sep 2009, 15:36
The other way to look at it is to simply have 5 donuts, with two dividers. You have 7 slots:
_ _ _ _ _ _ _
In those slots, you can either put a donut, O, or a divider, . The number preceding the first divider goes to Person 1, the number between the two dividers goes to Person 2, and the number after the 2nd divider goes to Person 3.
For example:
O O  O O  O   O O O O O  O  O O O O
etc.
You choose 2 spots out of the 7 to put the dividers, hence the answer is 7C2.



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Re: Larry, Michael, and Doug have five donuts to share. If any one of the [#permalink]
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19 Jul 2011, 05:35
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Here is another intuitive way, but certainly the formula will help in bigger calculations.
to get the answer, see how can we get sum "5" with 3 numbers.
1) 0,0,5 = 3 combinations or 3! /2!
2) 0,1,4 = 6 combinations or 3!
similarly 3) 0,2,3 = 6 cobinations 4) 1,1,3 = 3 combination 5) 2,2,1 = 3 combination
total =21



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Re: Larry, Michael, and Doug have five donuts to share. If any one of the [#permalink]
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19 Jul 2011, 06:01
bhushan252 wrote: OA A
This is little trivial rule in combinations
The formula for distributing n identical things amongst r people such that any person might get any number of things ( including 0) is (n+r1)C(r1)
in this problem n=5 & r=3 hence 7C2 = 21 ways
**PS : if you want no of choices excluding 0 then => (n1)C(r1) = 4C2 = 6 ways dear bhushan got your formula for cases when assigning 0 is allowed please explain the formula for cases where assigning of 0 is not allowed for example the question below If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible is it as simple as it looks here is my understanding 9 total digits out of which 3 places to assign we cant assign 0 so the total no of ways will be (91) C (31)= 8 C 2 =28 is it correct
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