exactly as it is in the photo

The 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I have a few questions about the 2nd part of the question, the one that asks for the number of ways to distribute $n$ indistinguishable objects into $k$ distinguishable boxes, where each box can contain at most $m$ objects.
I think the answer is $\binom{n+k-1}{k}$, but I'm not sure how to prove it. I know that the number of ways to distribute $n$ distinguishable objects into $k$ distinguishable boxes is $k^n$, but I'm not sure how to use that here.

Biko03 2019-02-25: Let us consider the number of ways to distribute the $n$ objects into the $k$ boxes such that no box is empty.
To do this, we can use the principle of inclusion exclusion. We have $k$ choices for the first object, $k-1$ choices for the second, $k-2$ choices for the third, etc... and so on. This gives us $k(k-1)(k-2)\cdots (k-n+1)$ ways to distribute the objects into the boxes. However, this counts the number of ways to distribute the objects without regard to the order in which they are placed into the boxes. Since the order in which the objects are placed does not matter, we must divide by $n!$ to correct for this overcounting. Thus, the number of ways to distribute the objects is
$$\frac{k(k-1)(k-2)\cdots (k-n+1)}{n!} = \frac{k!}{(k-n)!n!}$$
The number of ways to distribute the objects such that each box receives at least one object is then
$$\frac{k!}{(k-n)!n!}-\frac{k!}{(k-n-1)!n!}$$
This simplifies to
$$\frac{k!}{(k-n)!n!}-\frac{k!}{(k-n-1)!n!} = \frac{k!}{(k-n)!n!}\left(1-\frac{n}{k-n-1}\right) = \frac{k!}{(k-n)!n!}\left(\frac{k-n-1}{k-n-1}\right) = \frac{k!}{(k-n)!n!}$$
Thus, the number of ways to distribute $n$ indistinguishable balls into $k$ distinguishable boxes such that each box contains at least one ball is $\boxed{\binom{k}{n}}$.

# Answer

\binom{k}{n}$$

Very nice!!!! Perfect for someone addicted to Wanda like myself