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Counting strikes me again...HELP!

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Counting strikes me again...HELP!

by cypherskull » Sat Aug 25, 2012 11:43 am
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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Sunit

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by Brent@GMATPrepNow » Sat Aug 25, 2012 11:59 am
cypherskull wrote: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
This question is probably out of scope for the GMAT.
Nevertheless, here's my solution.
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Answer = A

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Brent
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by GaneshMalkar » Sat Aug 25, 2012 11:04 pm
cypherskull wrote: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

There is a very simple tool to tackle this....

If we want to distribute n identical things among r groups such that each gets a any number of things(including 0) then the no of ways in which it can be done is n+r-1Cr-1

So applying this n = 5; r = 3
then (5+3-1)C(3-1) = 7C2 = 21
If you cant explain it simply you dont understand it well enough!!!
- Genius
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