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

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Director
Joined: 07 Jun 2004
Posts: 614
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Kudos [?]: 335 [2] , given: 22

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

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Question Stats:

53% (02:19) correct 47% (01:31) wrong based on 252 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
[Reveal] Spoiler: OA

_________________

If the Q jogged your mind do Kudos me : )

Math Expert
Joined: 02 Sep 2009
Posts: 28784
Followers: 4605

Kudos [?]: 47682 [9] , given: 7130

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

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
_________________
Director
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Joined: 04 Jan 2011
Posts: 837
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Schools: Haas '18, Kelley '18
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GMAT 3: 750 Q51 V40
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WE: Education (Education)
Followers: 39

Kudos [?]: 157 [0], given: 78

Re: Combinations tough [#permalink]  04 Feb 2011, 17:46
Manager
Joined: 02 Jan 2013
Posts: 57
GMAT 1: 750 Q51 V40
GPA: 3.2
WE: Consulting (Consulting)
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Kudos [?]: 36 [0], given: 2

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
Manager
Joined: 18 Oct 2011
Posts: 91
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Followers: 2

Kudos [?]: 40 [0], given: 0

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
Manager
Joined: 06 Feb 2013
Posts: 60
Followers: 1

Kudos [?]: 21 [0], given: 35

Re: Combinations tough [#permalink]  03 Sep 2013, 07:05
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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There are times when I do not mind kudos...I do enjoy giving some for help

Manager
Joined: 23 May 2013
Posts: 103
Concentration: Strategy, Social Entrepreneurship
GMAT Date: 08-05-2015
Followers: 1

Kudos [?]: 49 [0], given: 29

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.

{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}
Intern
Joined: 12 Sep 2012
Posts: 25
GMAT 1: 550 Q49 V17
Followers: 0

Kudos [?]: 1 [0], given: 14

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.

Math Expert
Joined: 02 Sep 2009
Posts: 28784
Followers: 4605

Kudos [?]: 47682 [1] , given: 7130

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]  05 Jul 2014, 06:09
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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.
_________________
Intern
Joined: 18 Jun 2014
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 274

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]  06 Jul 2014, 04:50
Why 3^5 is not a correct answer?
Considering each donut has 3 possibilities(L,M &D)...
Kindly regards
Math Expert
Joined: 02 Sep 2009
Posts: 28784
Followers: 4605

Kudos [?]: 47682 [0], given: 7130

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]  06 Jul 2014, 09:29
Expert's post
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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.
_________________
Intern
Joined: 14 Dec 2010
Posts: 2
Schools: ISB
Followers: 0

Kudos [?]: 0 [0], given: 5

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
Director
Joined: 03 Feb 2013
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Concentration: Operations, Strategy
GMAT 1: 760 Q49 V44
GPA: 3.88
WE: Engineering (Computer Software)
Followers: 59

Kudos [?]: 402 [0], given: 533

Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]  20 Dec 2014, 09:53
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] 20 Dec 2014, 09:53
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