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

1 Answer
Jun 2, 2017

Answer:

#(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#