# Yesterday I posed the following question:
### How many ways can you place 20 identical balls into 4 unique bins?
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The answer is actually pretty cool and easy if you think about it in a certain light.
Let '0' represent a ball and '1' represent a barrier between 4 bins. (Therefore there are 3 barriers):
```
00100010000000001000000
```
This is just one possible way to place 20 balls in 4 bins, but it turns out that the exact number of ways is just the number of bitstrings of length 23 which contain 3 ones!
We "choose" three of the 23 bits to be barriers and the rest to be balls.

### This is C(23,3) = 23!/(20!3!) = 23\*22\*21/6 = 1171

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In my opinion, this is the most interesting of the basic counting arguments in combinatorics!