Suppose 4 kids have boxes and there are 5 balls we plan to put in the boxes. How many ways can we put balls in the boxes?u

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- Nov 8th 2007, 01:43 PMchadlyterCombination or permutation
Suppose 4 kids have boxes and there are 5 balls we plan to put in the boxes. How many ways can we put balls in the boxes?u

- Nov 8th 2007, 01:59 PMgalactus
If I am understanding correctly. You are placing 5 identical balls into 4 different boxes?.

There are C(5+4-1,5)=C(8,5) ways to do that.

If the balls are distinct, then 4^5 ways. - Nov 9th 2007, 04:12 AMchadlyterAlways true?
Will this always be true, like (n + m -1, n) and I have another problem that reads like this so I think it is 18 choose 15 but not quite sure, FInd the number of ways of assigning 15 distinct paintings to 18 different dormitories so that no dormitory receives more than one painting.

- Nov 9th 2007, 07:49 AMPlato
No, there is no such generalization as ‘always true’!

$\displaystyle {{m+n-1} \choose m}$ is the number of ways to put m identical object into n different cells.

The number of ways of assigning 15 distinct paintings to 18 different dormitories so that no dormitory receives more than one painting is the count of the number of one-to-one functions from a set of 15 to a set of 18: $\displaystyle P(18,15) = \frac{{18!}}{{3!}}$. - Nov 13th 2007, 10:01 AMchadlyterExplanation
Why does this m + n -1, m work and what does the minus one represent. I just would like to understand the meaning behind all of this working?

- Nov 13th 2007, 10:15 AMPlato
Here you can read about it.