Distributing Identical Objects To Distinct Bins

There are $n$ tennis balls all identical to each other, and $m$ tennis courts each distinct, let's say marked $1\ ...\ m$

In how many ways can we distribute these tennis balls amongst to the tennis courts without any restrictions, which means one or more tennis courts may receive $0$ tennis balls?

The approach is easy and we will see that by assigning numbers to $m$ and $n$. Let us say there are $5$ tennis ball and $3$ tennis courts.

Let us write mark each ball with the letter $b$ and arrange them in a line. They would look as follows:

$\enclose{circle}B\ \enclose{circle}B\ \enclose{circle}B\ \enclose{circle}B\ \enclose{circle}B$

Now, if we take $2$ partition sticks and place them randomly between the balls, as shown below:

$\enclose{circle}B\ \left|\ \enclose{circle}B\ \enclose{circle}B\ \right|\ \enclose{circle}B\ \enclose{circle}B$

The balls to left of the first partition stick ($1$ ball) goes to court $1$

The balls in between the first and the second partition sticks ($2$ balls) go to court $2$

The balls to the right of the second partition stick ($2$ balls) go to court $3$

Observe that if we place the first partition stick to the left of the first ball as shown below:

$\left|\ \enclose{circle}B\ \enclose{circle}B\ \enclose{circle}B\ \right|\ \enclose{circle}B\ \enclose{circle}B$

Similarly if the two sticks are placed side by side then court $2$ receives $0$ balls, and if the right stick is placed after the last ball court $3$ receives $0$ ball.

Let us say, we have $7$ empty slots as shown below:

$\large{\underline{ }\ \underline{ }\ \underline{ }\ \underline{ }\ \underline{ }\ \underline{ }\ \underline{ }}$

and choose any $2$ slots (as shown below) for the partition sticks, we can fill up the remaining $5$ slots with the balls.

$\large{\underline{ }\ \underline{ }\ \underline{|}\ \underline{ }\ \underline{|}\ \underline{ }\ \underline{ }}$

Since the balls are identical, the number of ways we can arrange the $5$ balls is $1$.

Therefore, every distinct choice of slots for the position sticks will give us a unique way to distribute the tennis balls.

We can choose two positions for the partition sticks in $\xacomb{7}{2}$ ways.

Therefore, for $n$ tennis balls and $m$ courts, we need $n$ slots for the balls and $m-1$ slots for the partition sticks, hence a total of $n + m - 1$ empty slots, out of which we need to select $m-1$ slots for the partition sticks.

The number of ways to do this, which is the same as distributing $n$ identical objects to $m$ distinct bins, is: