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Larry, Michael, and Doug have five donuts to share. If any [#permalink]
04 Feb 2011, 15:05

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Difficulty:

75% (hard)

Question Stats:

53% (02:18) correct
47% (01:32) wrong based on 243 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?

Re: Combinations tough [#permalink]
04 Feb 2011, 15:20

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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: \({n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21\).

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
05 Jul 2014, 06:09

1

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deya wrote:

Dear Bunuel,

Thanks for the great explanation. But there is a catch that is not clear to me. I read all other similar types of questions and also went through the explanations given by you. There you have used 3 separators for all the cases but here you have used 2. Could you tell me how would I know that how many separators should I use.

Thanks in advance.

Distributing between 4 use 3 separators; Distributing between 3 use 2 separators. _________________

Larry, Michael, and Doug have five donuts to share. If any one o [#permalink]
17 Jan 2013, 08:17

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

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
17 Jan 2013, 10:56

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

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: \({n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21\).

Bunuel - could you explain how this problem would sound if I used a simple counting principle like Larry can get 5 doughnuts Michael 4...all the way to have 5! and therefore I would get 5!= 120, which would obviously be too easy but I am a little confused as to the difference in wording. Solution and those dividers make good sense, so thanks for that. _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
05 Mar 2014, 10:51

If you get confused about combinations, there's a simple way to count these combinations as well, by counting the number of ways 5 can be summed with 3 numbers.

Tip: For each set, we only have to consider numbers less than the first; for instance, we wouldn't consider {2,3,0} because that's already accounted for in a permutation of {3,2,0}

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
03 Jul 2014, 08:07

Dear Bunuel,

Thanks for the great explanation. But there is a catch that is not clear to me. I read all other similar types of questions and also went through the explanations given by you. There you have used 3 separators for all the cases but here you have used 2. Could you tell me how would I know that how many separators should I use.

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
19 Nov 2014, 08:59

Keep in mind that 2 people might not even get any donuts at all.. so total 7! and 5 donuts are identical and 2 not given are identical .. so 7!/ 5! 2! = 21

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