There are three objects of type A, two objects type B and one object of type C. In how many ways can the objects be arranged. The objects of each type are identical. We cannot tell each A apart, so any reordering of the objects of type A counts as the same arrangement.

The six objects can be arranged in 6! different ways. If the A objects were not identical, we could arrange them in 3! ways. Since they ARE identical, we must divide 6! by 3! to take account of this.

If the B objects were not identical, we could arrange them in 2! ways. Since they ARE identical, we must divide 6! by 2! to take account of this.

There is only one object of type C, and we can arrange it in 1!=1 way.

The total number of arrangements is

\[\frac{6!}{3! \times 2! \times 1!}\]

ways.
In general, if we have \[n\]

objects with \[n_1\]

objects of type 1, \[n_2\]

objects of type 2 ,..., \[n_k\]

objects of type \[k\]

with \[n_1 + n_2 +...+n_k =n\]

then the number of arrangements of the objects is \[\frac{n!}{n_1! n_2! ...n_k!}\]

.