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]

19 May 2006, 23:22

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

In general this problem can take multiple forms
1. how many ways can u separate N numbered balls (1..N) into X buckets
2. How many ways can u separate N identical balls into X buckets.
3. How many ways can u add X numbers to get N as the sum where
a) 1 or more numbers can be 0
b) Each number is > 0.

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20 May 2006, 04:52

The simplest aproach for this kind of problem that I am aware of would be to add (X -1) separators to the bunch of Y objects that you want to distribute among X people.

For instance, in your current question, the permutation that looks like OO|O|OO means that Larry got 2 , Michael got 1 and Doug got 2 donuts.

Hence the formula: (Y + X - 1 )! / ( Y! * (X-1)! )
You have to divide by ( Y! * (X-1)! ) since the separators and distributed objects are identical.

So the answer is 7!/(2!*5!) = 21

1) This problem is different, so its approach is different as well. Since each
ball can go to one of X baskets, total number of distributions will be:
X^N

2) This one is a general case of the problem explained in the beginning of this post.

3) I am not sure I understood this question. Please explain.

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20 May 2006, 08:05

deowl wrote:

3) I am not sure I understood this question. Please explain.

Should have been obvious.

Anyway.

3a) How many ways can we get distinct non-negative integers (x1,x2,x3,x4...xr) such that x1+x2+x3..xr = N where N is another distinct non-negative number.

Solved the same way as separating N identical balls into R urns.

b) Here we just add the condition that x1,x2,x3 are all greater than 0.

We can again solve this as the urns+balls problem i.e the number of ways of placing N balls into R urns with each urn having at least one ball. Keep the N balls in a row. We get N-1 spots between the balls which is a demarcator. We'll now choose (R-1) spots from the N-1 spots.

Answer is (n-1)C(r-1) i.e (n-1)!/(n-r)!(r-1)!

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Re: Worth revising once [#permalink]

20 May 2006, 10:28

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saha wrote:

(1) 5-0-0 in 3!/2! or 3 ways.
(2) 4-1-0 in 3! or 6 ways
(3) 3-1-1 in 3!/2 or 3 ways
(4) 3-2-0 in 3! or 6 ways
(5) 2-2-1 in 3!/2 or 3 ways

21 ways..

Re: Worth revising once [#permalink] 20 May 2006, 10:28

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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

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

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