# Problem C.1

## Balls and bins

In this problem, we investigate the effect of various assumptions on the number of ways of placing $n$ balls into $b$ distinct bins.

1. Suppose that the $n$ balls are distinct and their order within a bin does not matter. Argue that the number of ways of placing the balls in the bins is $b^n$.
2. Suppose that the balls are distinct and that the balls in each bin are ordered. Prove that there are exactly $(b + n - 1)!/(b - 1)!$ ways to place the balls in the bins. (Hint: Consider the number of ways of arranging $n$ distinct balls and $b-1$ indistinguishable stricks in a row).
3. Suppose that the balls are identical, and hence their order within a bin does not matter. Show that the number of ways of placing the balls in the bins is $\binom{b+n-1}{n}$. (Hint: Of the arrangements in part (b), how many are repeated if the balls are made identical?)
4. Suppose that the balls are identical and that no bin may contain more than one balls, so that $n \le b$. Show that the number of ways of placing the balls is $\binom{b}{n}$.
5. Suppose that the balls are identical and that no bin may be left empty. Assuming that $n \ge b$, show that the number of ways of placing the balls is $\binom{n-1}{b-1}$.

### Distinct balls, unordered

There are $b$ ways to place the first ball, then $b$ ways to place the second and so on. There are $n$ balls, so the total number of ways is $b^n$.

### Distinct balls, ordered

As the hint indicates, this is isomporhic to arranging $n + b - 1$ items, out of which $b - 1$ are separators. The balls before the first separator go in the first bin, those between the first and the second go in the second bin, etc. There are $(n + b - 1)!$ ways to do that, but since the order of the separators does not matter, $(b - 1)!$ out of those are duplicated. Thus the answer is $(b + n - 1)!/(b - 1)!$.

### Identical balls, unordered

There are $(b + n - 1)!/(b - 1)!$ ways if the balls are distinct. If they are made identical, $n!$ of the arrangements are repeated for each position of the separators. We get $\frac{(b + n - 1)!}{n!(b - 1)!} = \binom{b + n - 1}{n}$ arrangements.

### Identical balls, max 1 per bin

This is reduced to selecting $n$ of the $b$ bins to put balls in, which is the definition of binomial coefficients - $\binom{b}{n}$.

### Identical balls, no bin left empty

This is fun. First, we put a ball in each bin and we're left with $n - b$ balls to put in $b$ bins. Now lets use part (c) - substituting $n - b$ for $n$, we get:

$$\binom{b + n - b - 1}{n - b} = \binom{n - 1}{n - b} = \binom{n - 1}{n - 1 - n + b} = \binom{n - 1}{b - 1}$$