Permutations with repetition

The idea is to first count as if all objects were different, which gives \(n!\) arrangements, and then correct the excess. Swapping two identical letters does not produce a new word, so each real arrangement has been counted \(n_1!\) times because of the internal permutations of the first type, \(n_2!\) times for the second, and so on. Dividing by \(n_1!\,n_2!\cdots n_k!\) undoes that multiplicity.

A guiding example is the word BANANA: its six letters have A three times and N twice, so there are \(\dfrac{6!}{3!\,2!\,1!}=60\) distinct arrangements, not \(6!=720\). The same reasoning counts binary strings with a fixed number of ones and paths in a grid, which are arrangements of repeated steps such as right or up.

Types that appear only once may be included or left out, since \(1!=1\); what must hold is that \(n_1+n_2+\cdots+n_k=n\). With only two types the formula agrees with \(\dfrac{n!}{n_1!\,n_2!}\), the binomial coefficient you will see later; with more types it is called a multinomial coefficient.

Typical mistakes: dividing by \(n_1+n_2+\cdots\) instead of by the product of the factorials, or using \(n!\) without correcting. When there are restrictions, such as fixing a letter at the start, set that letter aside and apply the formula to the rest.