# Thread: [SOLVED] distributing 'N' pennies to &quot;K&quot; children.

1. ## [SOLVED] distributing 'N' pennies to &quot;K&quot; children.

Can anybody through a light on this? I am not able toget the answer.

In how many ways can we distribute "N" coins to 'K" children so that everyone gets atleast one.

should it be n!/k! ?
if not, why?

2. Originally Posted by baz
Can anybody through a light on this? I am not able toget the answer.
In how many ways can we distribute "N" coins to 'K" children so that everyone gets atleast one.
We assume that the coins are identical.
The number of ways to place N identical objects into K distinct cells is $\displaystyle \binom{N+K-1}{N}$.
But the number of ways to do it so that no cell is empty equals
$\displaystyle \binom{(N-K)+(K-1)}{N-K}=\binom{N-1}{N-K}$.

3. Originally Posted by Plato
We assume that the coins are identical.
The number of ways to place N identical objects into K distinct cells is $\displaystyle \binom{N+K-1}{N}$.
But the number of ways to do it so that no cell is empty equals
$\displaystyle \binom{(N-K)+(K-1)}{N-K}=\binom{N-1}{N-K}$.
I dont get it that how the number of ways to place N identical objects in K distinct cells is $\displaystyle \binom{N+K-1}{N}$?

I dont get it that how the number of ways to place N identical objects in K distinct cells is $\displaystyle \binom{N+K-1}{N}$?