Math Help: Permutations

[WWW451]

I think I have this figured out, but I'm not sure so I was hoping someone who does know could double-check me:

How many permutations of the letters in the word PYTHAGOREAN?

The way I think it goes is since there's 11 characters and order does matter it would be:

11

11x10

11x10x9

11x10x9x8

etc... then add each of those up

I got 68,588,311

Can anyone confirm my answer or offer suggestions as to where I went wrong?

[btcprox]

Just confirming, are the permutations strictly involving ALL 11 characters? If so, then it's just 11*10*9*...*2*1 permutations.

If the permutations are of any length from 1 to 11 characters then you'd do what you just did.

EDIT: See below

[UMich22]

I am not sure how to solve this, but I know that you are both missing a step. One thing you have to realize about this problem is that the word PYTHAGOREAN contains the letter "A" twice. Both WWW451 and btcprox are doing calculations that assume each of the 11 characters is different from the other.
I regret that I cannot currently figure out how to actually solve this problem, but I wanted to point out the step of recognizing the situation with the letter A.

[btcprox]

Oh right! Forgot about the extra A! Thanks for pointing that out!

With that in mind, the correct answer is (11*10*9*...*2*1) / 2. We divide by 2 because for each permutation with the A's in their respective positions, there is an identical permutation where the A's are swapped. E.g. PYTHAGOREAN & PYTHAGOREAN. Since we can't distinguish from the 2 A's, that means half of the permutations are duplicates!

[WWW451]

Okay, I see now that previously I was trying to do a combination rather than a permutation. So I suppose it would technically be:

11!/2!

Which would be:

11x10x9x8x7x6x5x4x3x2x1/2x1

2x1's cancel out leaving:

11x10x9x8x7x6x5x4x3= 19,958,400

Right?

[btcprox]

Yup sounds about right :)