Probability Q

[rocknroll]

i found this online. why are they mulitplying by 3! at the end though? I get it but not the multiplying by 3! part? thanks in advance for your help

Question:
Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

Explanatory Answer
There are three possible cases that will satisfy the condition of forming three letter passwords with at least 1 symmteric letter.

Case 1: 1 symmetric and 2 asymmetric or
Case 2: 2 symmetric and 1 asymmetric or
Case 3: all 3 symmetric

= {(11C1 * 15C2) + (11C2 * 15C1) + 11C3} * 3!

= {11 * + * 15 + } x 6

= {1155 + 825 + 165} * 6

= 2145 * 6 = 12870

[DanaJ]

Thar 3! from the end comes from the fact that you have 6 possible ways (or 3!) to arrange any 3 given letters. Here's an example:

A B C
A C B
B C A
B A C
C A B
C B A

It's because of these alternative arrangement that you need to add that 3! to the end.

[rossmj]

I believe that the 3! is to account for the symmetric letter fitting into any of the three spots.

BAC
BCA

Are two different passwords

[m&m]

Question is AT LEAST 1 symmetric. So this includes total combinations of non-repeated 3 letter passwords - no symetrical

26*25*23 - 15*14*13
= 13*5*2(1*5*23 - 3*7*1)
= 130*(115-21)
=130*(94)
=9400 + 2820
=12,220

is that right?