Can anyone help with this problem?

Find the number of ways of arranging the letters A, A, B, B, C, C, so that the A's are not adjacent, the B's are not adjacent, and the C's are not adjacent.

Guest May 5, 2020

#1**0 **

A, A, B, B, C, C =6! / 2!.2!.2! = 90 permutations.

Each permutation will begin with one of letters in: [90 / 6] x 2 =30 cases for each of the 3 letters. And in 4/6 of the cases at least two of the same letters will be next to each other. And in 1/3 of the cases, they will not be. So, for each permutation beginning with A, you will have:

1/3 x 30 = 10 permutations when they will NOT be next to each other.

So, you will have a total of: **10 x 3 = 30 permutations** where the 3 letters will be separated from each other. Here are just 10 of them beginning with the letter A:

{A, B, A, C, B, C}, {A, B, C, A, B, C}, {A, B, C, A, C, B}, {A, B, C, B, A, C}, {A, B, C, B, C, A},{A, C, A, B, C, B}, {A, C, B, A, B, C}, {A, C, B, A, C, B}, {A, C, B, C, A, B}, {A, C, B, C, B, A} = 10 permutations.

That is it, unless I made mistake somewhere!.

Guest May 5, 2020