# Thread: Distinguishable permutations of AAABBCC. Help!

1. ## Distinguishable permutations of AAABBCC. Help!

Find the number of distinguishable permutations of the given letters "AAABBCC''.

First I thought I could just do this :

A1A2A3B1B2C1C2
""""""""""""" C2C1
"""""""""B2B1C1C2
""""""""""""" C2C1
A1A3A2B1B2C1C2...
so 2 start with A1, 2 with A2, and 2 with A2
then for every set of A there are two sets of B
and for every set of B there are two more sets of C
But apparently it's wrong.
Can anyone tell me why and more importantly where does this formula 7!/(3!2!2!) come from.

2. ## Re: Distinguishable permutations of AAABBCC. Help!

Originally Posted by SillyMath
Find the number of distinguishable permutations of the given letters "AAABBCC''.
First I thought I could just do this :

A1A2A3B1B2C1C2
""""""""""""" C2C1
"""""""""B2B1C1C2
""""""""""""" C2C1 if
A1A3A2B1B2C1C2...
so 2 start with A1, 2 with A2, and 2 with A2
then for every set of A there are two sets of B
and for every set of B there are two more sets of C
But apparently it's wrong.
Can anyone tell me why and more importantly where does this formula 7!/(3!2!2!) come from.
In North America this is known as the Mississippi problem
Consider "MISSISSIPPI" Put subscripts an the repeated letters: $MI_1S_1S_2I_2S_3S_4I_3P_1P_2I_4$
There are $11!$ ways to rearrange those eleven distinct letters.
Now we drop the subscripts we must divide by $2!\cdot 4!\cdot 4!$.
Thus the number of ways to rearrange the word MISSISSIPPI is $\frac{11!}{2!\cdot 4!\cdot 4!}$
You can generalize that.

3. ## Re: Distinguishable permutations of AAABBCC. Help!

Oh so I'm right! Good! Silly webworks! Thanx!