How many ways are there to arrange these four letters (A, B, C and D)? Using permutations, we know that there are P(4,3) (= 24) ways to do it:
Now lets say that there are two same letters and two unique letters (A, A, B and C). Now how many ways are there of arranging them? If you were to treat the two repeated letters as two unique letters, the arrangements would be the same as above:
There are 4! to arrange 4 letters, but since 2 letters are repeated, all the arrangements will appear twice:
The total number of arrangement is (4!/2 =) 12. Now lets try again but with three letters repeated (A, A, A and B):
Since there 3 repeated letters, then each arrangement will be repeated 3! times. For example, consider the arrangement [B A A A], there are 3! of arranging the three A's, so the same arrangement will be repeated 3! times. The total number of unique arrangements is (4!/3! =) 4.
You can probably assume that if you want to find the number of arrangements of n letters, where k letters are repeated, then there will n!/k! unique arrangements. Now lets say that my four letters are (A, A, B and B).
There are 4! ways to arrange 4 letters if we consider the first A and the second A to be unique, and also the first B and the second B to be unique, but if we consider two A's to be same, then there are (4!/2! =) 12 ways to arrange these letters, and if consider the 2 B's to be unique, then there are (12/2! =) 6 ways to arrange the letters.
Similarly, if there were 6 letters, with two A's and three B's (A, A, B, B, B, C), then the number of ways to arrange the letters is (6!/(2!*3!) =) 60.
We can generalize this and say that the number of arrangements of n objects with n1 identical objects, n2 identical objects, ..., nk identical objects:
