Number of ways to partition a set into subsets of given cardinality

Question

Set A contains 30 elements.

Set A = {1,2,3, ... ,29,30}

This set must be partitioned into 4 sets. One set must have cardinality 9, another cardinality 8, another cardinality 7, and the final one has cardinality 6.

|B| = 9
|C| = 8
|D| = 7
|E| = 6

How many ways can this be done? I think the answer is:

30! / (9!8!7!6!)

Answer 1

We could order the numbers and say that the first $9$ are meant to be the elements of $B$, the folllowing $8$ the elements of $C$, et cetera.

There are $30!$ possible orders.

Now look at some fixed result and realize that $9!8!7!6!$ orders of the $30!$ orders lead to that result.

This multiple counting can be repaired by dividing.

Answer 2

drhab gives one explanation; here’s another. There are $\binom{30}9$ ways to choose the $9$ members of the first set. Once that’s been done, there are $21$ numbers left, so there are $\binom{21}8$ ways to choose the $8$ members of the second set. That leaves $13$ numbers, so there are $\binom{13}7$ ways to choose the $7$ members of the third set, and the $6$ numbers that remain will be the fourth set. These three choices can be made in

$$\begin{align*} \binom{30}9\binom{21}8\binom{13}7&=\frac{30!}{9!\color{red}{21!}}\cdot\frac{\color{red}{21!}}{8!\color{blue}{13!}}\cdot\frac{\color{blue}{13!}}{7!6!}\\ &=\frac{30!}{9!8!7!6!} \end{align*}$$

different ways.