[Math] Number of ways to partition a set into subsets of given cardinality

combinatoricsset-partition

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!)

Best Answer

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.

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