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Larry, Michael, and Doug have five donuts to share. If any [#permalink]
 04 Feb 2011, 16:05
Question Stats:
60% (02:30) correct
40% (01:19) wrong based on 30 sessions
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: Combinations tough [#permalink]
04 Feb 2011, 16:20
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rxs0005 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 Consider this: we have 5 donuts d and 2 separators |, like: ddddd||. How many permutations (arrangements) of these symbols are possible? Total of 7 symbols (5+2=7), out of which 5 d's and 2 |'s are identical, so \frac{7!}{5!2!}=21. We'll get combinations like: dd|d|dd this would mean that Larry got 2 donuts, Michael got 1 donut and Doug got 2 donuts, so to the left of the first separator are Larry's donuts, between the separators are Michael's donuts and to the right of the second separator are Doug's donuts Answer: A. This can be done with direct formula as well: The total number of ways of dividing n identical items (5 donuts in our case) among r persons or objects (3 persons in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is {n+r-1}_C_{r-1}. In our case we'll get: {n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21. Similar question: integers-less-than-85291.html#p710836
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Re: Combinations tough [#permalink]
04 Feb 2011, 18:46
Nice! Thanks man 
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Larry, Michael, and Doug have five donuts to share. If any one o [#permalink]
17 Jan 2013, 09:17
Another awesome variation to this question (the one mentioned above) would be: How many integer solutions (x,y,z) are there to the equation: x+y+z = 20, where x is at least equal to 3, y is at least equal to 4, and z is at least equal to 5
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
17 Jan 2013, 11:56
should it not state identical donuts to share? relatively straightforward question but got confused as to whether we needed to use combinations formula and then arrange between the 3 people
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Re: Larry, Michael, and Doug have five donuts to share. If any
[#permalink]
17 Jan 2013, 11:56
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