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# Larry, Michael, and Doug have five donuts to share. If any

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Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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04 Feb 2011, 16:05
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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
[Reveal] Spoiler: OA

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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

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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04 Feb 2011, 18:46

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Larry, Michael, and Doug have five donuts to share. If any one o [#permalink]

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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]

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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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03 Sep 2013, 08:05
Bunuel wrote:
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

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

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.
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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05 Mar 2014, 11:51
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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.

{5,0,0} = 3 possibilities.
{4,1,0} = 6 possibilities.
{3,2,0} = 6 possibilities.
{3,1,1} = 3 possibilities.
{2,2,1} = 3 possibilities.
Total = 21 possibilities.

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}

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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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03 Jul 2014, 09:07
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.

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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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05 Jul 2014, 07:09
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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.

Distributing between 4 use 3 separators;
Distributing between 3 use 2 separators.
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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06 Jul 2014, 05:50
Why 3^5 is not a correct answer?
Considering each donut has 3 possibilities(L,M &D)...
Kindly regards

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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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06 Jul 2014, 10:29
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lolivaresfer wrote:
Why 3^5 is not a correct answer?
Considering each donut has 3 possibilities(L,M &D)...
Kindly regards

Because the donuts are not distinct.
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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19 Nov 2014, 09:59
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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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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20 Dec 2014, 10:53
L + M + D = 5
As L,M,D >= 0

We need to distribute 5 donots and 2 empty vessels. So 7!/(5!*2!) or 7C2 = 21 - A)
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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29 Dec 2015, 11:49
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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13 Jan 2016, 02:33
This problem similar to:

In how many ways can 3 distinct rings be worn on 4 fingers? (each finger can contain 0-3 rings)

OR

In how many ways can 3 distinct letters be distributed among 4 letterboxes? (each letterbox can contain 0-3 letters)

The solution to this problem is given as 4^3, as you have 3 slots (each slot representing a ring or a letter) and each slot can have 4 possibilities

4 4 4 = 4^3

Yet, the 3^5 does not work in the case of the donut. Please explain.

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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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08 Mar 2017, 22:42
Hello from the GMAT Club BumpBot!

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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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13 Jul 2017, 09:46
Can someone please explain this separator method? Or can anyone post a link where it is explained? I dont understand this solution

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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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16 Jul 2017, 17:37
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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

Let D be a donut, so we have DDDDD to distribute among three people. Also, we can use “|” as a separator, and we need two of them since there are three people. For example, D|DDD|D means Larry gets 1 donut, Michael 3, and Doug 1, and DDDDD|| means Larry gets 5 donuts, Michael 0, and Doug 0. Thus, the problem becomes how many ways we can arrange 5 Ds and 2 strokes. To solve it, we can use the formula for permutation of indistinguishable objects:

7!/(5! x 2!) = (7 x 6 x 5!)/(5! x 2) = 42/2 = 21

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Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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16 Jul 2017, 19:59
$$0+1+4$$ - this can be done in $$6$$ ways: $$014, 104,401,140,410,041$$

$$1+1+3$$ - this can be done in $$3$$ ways: $$\frac{3!}{2!}$$

$$1+2+2$$ - this can be done in $$3$$ ways: $$\frac{3!}{2!}$$

$$2+3+0$$ - this can be done in $$6$$ ways

$$0+0+5$$ - this can be done in $$3$$ ways

Total no. of ways = $$6+3+3+6+3 = 21.$$
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Larry, Michael, and Doug have five donuts to share. If any   [#permalink] 16 Jul 2017, 19:59
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