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

Jun 2, 2017

(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