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

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Senior Manager
Joined: 19 Jul 2006
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Larry, Michael, and Doug have five donuts to share. If any [#permalink]

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28 Dec 2006, 23:31
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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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VP
Joined: 25 Jun 2006
Posts: 1161

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28 Dec 2006, 23:48
i get A.

i got this by counting.

1) all 5 donuts go to 1 person. Possibility: 3.

2). all 5 donuts go to 2 ppl:
the one without anything: poss: 3.
the other 2 pp: have 4 ways.
So totally: 3*4 = 12.

3). all 5 have donuts: the way to divide can only be:
1, 2, 2,
1, 3, 1

but this arrange can rotate with the 3 people, so it become: 3!/2 + 3!/2 = 6. 3!/2 is needed because donuts are all same without difference. 3! is counting the same arrangement twice.

3 + 12 + 6 = 21.

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Senior Manager
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29 Dec 2006, 05:44
good approach, tennis ball.

It is always robust and safe to divide into cases and count, and it should solve any question in gmat in a reasonable time.

i have another method for solving "partitioning" questions. however - this method works well if everybody should get at least 1. so i'll rephrase: instead of 3 people sharing 5 doughnuts with possibility of getting none. let 3 people share 8 doughnuts each get at least one. think about it for a minute and see that it is essentially the same question.

here is my method to solve it:
place the 8 doughnuts in a line. put 2 markers between doughnuts in different places to split the doughnut line into three groups, and you got a partition of the 8 doughnuts to three. the leftmost goes to person1, the middle goes to person2 the right goes to person3.
so we are left to count how many ways there are to place two markers between 8 doughnuts.
there are 7 places between 8 doughnuts, and if we want to put 2 markers we have 7C2 = 7*6/2 = 21 ways to do so.

this of course can be generalized to any number of doughnuts and any number of persons (but it may not work for apples though.... )

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Manager
Joined: 18 Nov 2006
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29 Dec 2006, 13:45
AK 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

------------------------------------
A..

5,0,0 -> 3
4,1,0 -> 3*2
3,2,0 -> 3*2
3,1,1 -> 3
2,2,1 -> 3
Total =21

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Senior Manager
Joined: 08 Jun 2006
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31 Dec 2006, 17:36
Andr359 wrote:
Yep. A. You can see that a typical answer for this sort of questions is the sum of the series of integers from 1 to n+1 (n = number of items to distribute). In this case n = 5, thus the answer is 21. For n = 6 itÂ´ll be 28. For n = 7, 36. And so on and so forth.

Excellent observation .....
Can you explain why it holds this pattern.. [1 ...+ (n +1)]?

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Manager
Joined: 28 Aug 2006
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31 Dec 2006, 21:09
Because the formula (n+2)C2 is same as adding n + 1 numbers with first term as 1

(n+1){2*1 + (n+1-1)*1}/2 is the sum of arithmetic progression

Both give (n+2)(n+1)/2

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Senior Manager
Joined: 24 Nov 2006
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01 Jan 2007, 13:54
anindyat wrote:
Andr359 wrote:
Yep. A. You can see that a typical answer for this sort of questions is the sum of the series of integers from 1 to n+1 (n = number of items to distribute). In this case n = 5, thus the answer is 21. For n = 6 itÂ´ll be 28. For n = 7, 36. And so on and so forth.

Excellent observation .....
Can you explain why it holds this pattern.. [1 ...+ (n +1)]?

Actually, for the particular case in which the number of people to whom distribute the items is 3, that formula holds. In general, for n people and I items to distribute, the formula is (I+n-1)Cr(n-1), or (I+n-1)! / (I! * (n-1)!). We verify that it works for the problem: I=5, n=3: (5+3-1)Cr(2) = 7Cr2 = 21.

When n=3: (I+3-1)Cr(3-1) = (I+2)Cr2 = (I+2)*(I+1)/2 = formula for the sum of integers from 1 to I+1.

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01 Jan 2007, 13:54
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