# How many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes?

How many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes so that the boxes have one, two, three, four, and five objects in them ?

I tried to solve it as =>

$C(15,1) * C(14,2) * C(12,3) * C(9,4) * C(5,5)$ and

since placing total number of objects in each box matters here ,i.e, objects being placed in boxes in the count $1,2,3,4,5$ is different from $3,1,5,4,2$ and so on.

So, i multiplied it by $5!$

Hence, final answer I think should be

$5! * C(15,1) * C(14,2) * C(12,3) * C(9,4) * C(5,5)$

I don't have answer for this question. So, is my approach right ?

You counted the ways to partition the objects into groups of distinct sizes $1,2,3,4,5$, and then ways to arrange those groups into the boxes.   That is what you wanted to count, and how you could count it.
Also written as $5!\dbinom{15}{5,4,3,2,1}$ using the multinomial coefficient notation.
• Stirling's formula states: $\lim\limits_{n\to\infty}\dfrac{n!}{n^{n+1/2}~e^{-n}}=\sqrt{~2~\pi~}$, which is not at all useful. You *may* be thinking of Stirling Numbers of the Second Kind $\left\{\begin{matrix}n\\k\end{matrix}\right\}$, which count ways to partition a set of $n$ elements into $k$ non-empty subsets. Though, this is not useful either. Or are you thinking of a different "Sterling" all together? – Graham Kemp Nov 16 '16 at 3:10