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Halle, Julia and Drew have 5 donuts to share. If one of them

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Halle, Julia and Drew have 5 donuts to share. If one of them [#permalink] New post 13 Oct 2004, 21:04
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Halle, Julia and Drew have 5 donuts to share. If one of them can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed. Answer:21
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combination problem [#permalink] New post 13 Oct 2004, 21:33
these are the no of ways:

0 2 3 - 3! ways

2 2 1 - 3!/2 ways

5 0 0 - 3!/2 ways

4 1 0 - 3! ways

1 1 3 - 3!/2 ways

So total no of ways = 21
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 [#permalink] New post 14 Oct 2004, 02:13
Is there any tricky way which could be used if question is p donuts and n persons ?
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 [#permalink] New post 16 Oct 2004, 06:51
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Hmmm.

This is a combinations with repetition problem.

Just have to think about it the right way. Basically we have a bag of donuts from which we will pull 5 times. Any donut removed from the bag can be a Halle, Julia, or Drew Donut.

The general formula is:
C(n+r-1,r)
n number of elements
r-combinations.

So
C(5+3-1,5) = C(7,5) = C(7,2) = 21

The way that this makes sense is if we consider aligning the 5 donuts (*) and placing separators between them to indicate to whom they belong.
So,
*****||
would indicate that Halle has all the donuts,
|*****|
would indicate that Julia has all the donuts, and
||*****
would indicate that Drew has all the donuts.

So the answer is the number of ways to pick the 5 *'s or the 2 |'s out of the set of 7.

Thomas
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 [#permalink] New post 16 Oct 2004, 23:02
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tygel, this is a great logic

Extending this logic further can also be used to solve problems like these:

1. In how many ways can 20 be expressed as sum of 5 non negative integers.

using similar logic : you have 20 *s are 4 seperators '|' and u have to chose 4 '|' and u can do that in 24C4 ways.

TO add to this twist, in how many ways can the number 20 be written as a sum of 5 positive integers?
you have 20 *'s and u have to place 4 '|' in between them, i.e. no of ways in which 4 '|' can be placed in 19 places, where all '|' are identical. 19C4.

To summarise, no of ways in which n things can be divided among r persons so that each of them can receive 0 or more is n+r-1Cr-1. In tapsemi's Q. n= 5, r = 3, so reqd ways = 7C2 = 21.

and no of ways in which n things can be divided among r persons so that each of them receive at least 1 is n-1Cr-1
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 [#permalink] New post 17 Oct 2004, 10:09
This brings up an interesting question ...

If the question were ...

Given

x_1 + x_2 + x_3 = 6

How many positive solutions are there where x_1, x_2, and x_3 are all positive numbers.

Then the equation you gave works ...

(6-3+3-1)C2 => 5C2 => 10

We can see this is true by listing out the 10 solutions (x_1,x_2,x_3):

1, 1, 4
1, 4, 1
4, 1, 1
1, 2, 3
1, 3, 2
2, 1, 3
2, 3, 1
3, 1, 2
3, 2, 1
2, 2, 2

However, if the question were how many ways can 3 postive numbers be added to form 6 the answer is different.

There are only 3 unique solutions
1, 1, 4 (3 permutation)
1, 2, 3 (6 permutation)
2, 2, 2 (1 permutation)

Anyone care to take a stab at the general solution for that one?

Thomas
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 [#permalink] New post 19 Oct 2004, 07:10
Thank you target and tyagel for your inputs !
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Re: combination problem [#permalink] New post 17 Sep 2009, 15:27
Thanks!!!

What would be the case if the donuts are considered different from each other? Is this the same?
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Re: combination problem [#permalink] New post 02 May 2011, 08:24
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n things to be divided among 3 people so that each gets from 0-n = C(n+r-1, r-1) = 21
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Re: combination problem   [#permalink] 02 May 2011, 08:24
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