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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 one of the  [#permalink]

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New post 07 Aug 2009, 06:57
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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
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Re: Combinations tough  [#permalink]

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

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New post 07 Aug 2009, 07:51
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OA A

This is little trivial rule in combinations

The formula for distributing n identical things amongst r people such that any person might get any number of things ( including 0) is (n+r-1)C(r-1)

in this problem n=5 & r=3 hence 7C2 = 21 ways

**PS : if you want no of choices excluding 0 then => (n-1)C(r-1) = 4C2 = 6 ways
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Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

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New post 22 Sep 2009, 16:36
The other way to look at it is to simply have 5 donuts, with two dividers. You have 7 slots:

_ _ _ _ _ _ _

In those slots, you can either put a donut, O, or a divider, |. The number preceding the first divider goes to Person 1, the number between the two dividers goes to Person 2, and the number after the 2nd divider goes to Person 3.

For example:

O O | O O | O
| | O O O O O
| O | O O O O

etc.

You choose 2 spots out of the 7 to put the dividers, hence the answer is 7C2.
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Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

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New post 19 Jul 2011, 06:35
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Here is another intuitive way, but certainly the formula will help in bigger calculations.

to get the answer, see how can we get sum "5" with 3 numbers.

1) 0,0,5 = 3 combinations or 3! /2!

2) 0,1,4 = 6 combinations or 3!

similarly
3) 0,2,3 = 6 cobinations
4) 1,1,3 = 3 combination
5) 2,2,1 = 3 combination

total =21
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Re: Larry, Michael, and Doug have five donuts to share. If any  [#permalink]

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New post 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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New post 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.

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

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New post 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.

Thanks in advance. :-D


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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New post 06 Jul 2014, 05:50
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]

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

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New post 19 Nov 2014, 09:59
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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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New post 20 Dec 2014, 10:53
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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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New post 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

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

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

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New post 11 Jun 2018, 00:30
lbsgmat 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


Using the slot method, where D denotes the donut & l denotes the slot.

D D D D D l l

So # of ways of dividing the donuts is the # of possible arrangements of 5 D's & 2 l's = 7!/5!*2! = 21

Answer A.

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

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New post 03 Apr 2019, 19:27
Consider 5 donuts: O O O O O

Divide it among 3 people:
Case 1: L=0
1.1: L=0, M=6, D=0
1.2: L=0, M=5, D=1 ...
Thus we will be able to do this till L=0, D=0, L=6. Total = 6 possibilities

Case 2: L=1
L=1,M=5, D=0.
Similar to above we can do this for Total = 5 possibilities

Thus, we can continue such iterations till L=6

From first two cases we can simply observe that there is a series formed: 6+5+4+3+2+1

6+5+4+3+2+1 =21
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Re: Larry, Michael, and Doug have five donuts to share. If any   [#permalink] 03 Apr 2019, 19:27
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