# Combinatorics, permutations of letters

1. Dec 17, 2012

### Hannisch

1. The problem statement, all variables and given/known data
How many "words" with 5 letters can be created from the letters in the word ALGEBRA? Each letter can be used only once.

2. Relevant equations

3. The attempt at a solution
I know the answer to this (1320) and I know how I got to it (which I'll describe in a minute), but I know that while my way of getting it is right, it's. not the easiest, and I can't seem to figure out how on earth they want me to do it.

I know that choosing 5 things from a set with 7 elements can be done in

$\frac{7!}{(7-5)!}$

ways. I also know that if I were to use all the letters of ALGEBRA the answer would be

$\frac{7!}{2!}$

since one letter is used twice. In this case these happen to be exactly same, but I can't seem to put these principles together to get a correct answer. Help, please?

The way I solved it was more of a brute force solution. The correct answer is 11 * 5!, and I got to that conclusion by reasoning that 5! of the words are made by the letters LGEBR, which have no recurring letters. Then I "exchanged" one of the letters from that to an A, one at a time, and then to both As. This can be done in 11 ways:

LGEBR
AGEBR
LAEBR
LGABR
LGEAR
LGEBA
AAEBR
LAABR
LGAAR
LGEAA
AGEBA

And all of them can be chosen in 5! different combinations. I know this is the correct answer (it's an online based homework, I've inputted this and it says I'm right).

2. Dec 17, 2012

### tiny-tim

Hi Hannisch!

That certainly works.

But quicker would be to split the problem into three …

count separately the number of words with no As, with one A, and with 2As.

3. Dec 17, 2012

### haruspex

Slightly quicker still, just separate the cases "at most one A" (just as with all the other letters) and "2 As"

4. Dec 18, 2012

oh yes!