Question

Find the number of distinguishable permutations of:

2 Bs, 3 Hs, and 6 Os
(B, B, H, H, H, O, O, O, O, O, O)


If you could explain how / the formula, that would be great

Answer 1

There are a total of 2 + 3 + 6 = 11 letters, which gives rise to 11! possible permutations.

Consider one such permutation:

BBHHHOOOOOO

I write one of the Bs in the bold for emphasis. This permutation is not distinguishable (but for the temporary boldface) from

BBHHHOOOOOO

which is to say they both count as the same permutation. To avoid counting this twice, you would divide the total number of permutations by the number of ways you can permute the identical character. In the case of Bs, this can be done in 2! = 2 ways.

Similarly, the 3 Hs can be rearranged in 3! = 6 ways, and the 6 Os can be rearranged in 6! = 720 ways.

Then the total number of distinguishable permutations is

11! / (2! • 3! • 6!) = 4620

As a formula: given a word of length
`N` with
`n` different characters that respectively occur
`k_n` times, the total number of distinct permutations is

`(N!)/(k_1! k_2! k_3! \cdots k_n!)`

where
`k_1+k_2+k_3+\cdots+k_n=N`.

Try this with some simple examples:

• GREECE

This has length 6, with E occurring three times, and every other letter occurs once. The total number of distinct permutations of GREECE is

6! / (1! • 1! • 1! • 3!) = 120

• BANANA

This also has length 6, with B occurring once, N twice, and A three times, so the total number of distinct permutations of BANANA is

6! / (1! • 2! • 3!) = 60

• STATISTICALLY

This has length 13, with one each of C and Y; two each of S, A, I, L; and three Ts. Then the number of distinct permutations is

13! / (1! • 1! • 2! • 2! • 2! • 2! • 3!) = 64,864,800