We shall now determine the number of permutations of a set of n objects when certain of the objects are indistinguishable from each other. To set this straight in our minds, consider the following example. In general, the same reasoning shows that there are

$\displaystyle\frac{n!}{n_1! n_2! ... n_r!}$

different permutations of n objects, of which n1 are alike, n2 are alike, ..., are alike.

How many different letter arrangements can be formed using the letters P E P P E R?
We first note that there are 6! permutations of the letters P1 E1 P2 P3 E2 R when the 3 P's and the 2 E's are distinguished from each other. However, consider any one of these permutations -- for instance, P1 E1 P2 P3 E2 R. If we now permute the P's among themselves and the E's among themselves, then the resultant arrangement would still be of the form P E P P E R. That is, all 3!2! permutations

P_1 P_2 E_1 P_3 E_2 R & P_1 P_2 E_2 P_3 E_1 R \\
P_1 P_3 E_1...
...E_2 P_2 E_1 R \\
P_3 P_2 E_1 P_1 E_2 R & P_3 P_2 E_2 P_1 E_1 R \\

are of the form P E P P E R. Hence there are $\displaystyle\frac{6!}{3!2!}=60$ possible letter arrangements of the letters P E P P E R.         $\qed$