Question

How do you find the number of distinguishable permutations of the group of letters: A, A, G, E, E, E, M?

Answer 1

`(7!)/(2!3!) = 420`

Explanation:

Given:

`A, A, G, E, E, E, M`

For the sake of discussion, let's distinguish all of the letters by adding subscripts:

`A_1, A_2, G, E_1, E_2, E_3, M`

The number of distinguishable permutations of these marked letters is:

`7! = 7*6*5*4*3*2*1 = 5040`

If we now remove the marking, then some of these distinguishable permutations become indistinguishable.

In fact, each of the previously distinguishable arrangements has a total of `2!*3! = 12` variants, now indistinguishable.

So the total number of distinguishable permutations is:

`(7!)/(2!3!) = 5040/(2*6) = 420`