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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] New post 04 Feb 2011, 15:05
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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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Re: Combinations tough [#permalink] New post 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: \({n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21\).

Similar question: integers-less-than-85291.html#p710836
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Re: Combinations tough [#permalink] New post 04 Feb 2011, 17:46
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Larry, Michael, and Doug have five donuts to share. If any one o [#permalink] New post 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
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink] New post 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
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Re: Combinations tough [#permalink] New post 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

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


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

Thanks in advance. :-D
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink] New post 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.

Thanks in advance. :-D


Distributing between 4 use 3 separators;
Distributing between 3 use 2 separators.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink] New post 06 Jul 2014, 04:50
Why 3^5 is not a correct answer?
Considering each donut has 3 possibilities(L,M &D)...
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink] New post 06 Jul 2014, 09:29
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink] New post 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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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink] New post 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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