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Straight-forward: given $x$ identical balls and $n$ boxes, how many different ways can we place the balls into the boxes? A box can be empty.

example: x = 4, n = 3

|OOOO| - | | - | | |OOO | - |O | - | | |OOO | - | | - | O | |OO | - |OO | - | | |OO | - |O | - |O | |OO | - | | - |OO | |O | - |OOO | - | | |O | - |OO | - |O | |O | - |O | - |OO | |O | - | | - |OOO | | | - |OOOO| - | | | | - |OOO | - |O | | | - |OO | - |OO | | | - |O | - |OOO | | | - | | - |OOOO|

ANS = 15

In this case, seems like we have a sum $(1+2+3+4+5)$.

example: x = 3, n = 4

|OOO| - | | - | | - | | |OO | - |O | - | | - | | |OO | - | | - |O | - | | |OO | - | | - | | - |O | |O | - |OO | - | | - | | |O | - |O | - |O | - | | |O | - |O | - | | - |O | |O | - | | - |OO | - | | |O | - | | - |O | - |O | |O | - | | - | | - |OO | | | - |OOO| - | | - | | | | - |OO | - |O | - | | | | - |OO | - | | - |O | | | - |O | - |OO | - | | | | - |O | - |O | - |O | | | - |O | - | | - |OO | | | - | | - |OOO| - | | | | - | | - |OO | - |O | | | - | | - |O | - |OO | | | - | | - | | - |OOO|

ANS = 20

In this case, seems like we have a sum $(1+3+6+10)$.

Daniel
  • 732

1 Answers1

0

The way to place x balls in n boxes (where boxes can be empty) is:

$$\binom{x + n - 1}{n - 1}$$

Daniel
  • 732