# Thread: Permutation Problem

1. ## Permutation Problem

I was having an issue with this problem in my homework... was wondering if someone could help me out.

how many ways can u spell a 4 letter word using letters from CALCULATE using:
1) all different letters
2) two "a"s and another identical letter pair
3) two identical letter pairs
4) one pair of identical leters and 2 other different letters

and how many total 4 letter possibilities are there?

Thanks for your help!

2. Have you tried to do anything on this problem?
Here is a start on #4.
${3 \choose 1}{5 \choose 2} \frac {4!} {2!}$, Choose the pair, choose the two other letters, then scramble.

3. Thanks for the reply... well, this is what I haev so far:

1) All different letters, so you have 6 (CALUTE) letters to choose from to make 4 letter scrambels. So, the answer is P(6,4) = 6!/2! = 360 possibilities

2) Two "a"s, and another identical pair. So, I assumed you could only have two other "l"s, or two other "c"s. So, using AACC, P(4;2,2) = 4!/2!2! = 6. Same thing with AALL. So, there are 12 possibilities

3) With any two identical pairs. So, same as 2), but you add another 6 possibilities in the case of CCLL. So, 18 possibilities.

4) I'm not sure how to tackle this. So far in class, we've only learned permutation formulas, nothing about combinations (the choose thing). So... is there a way to solve this using permutations only?

Thanks for your help, and it'd be great if you could confirm my answers to 1), 2) and 3) are correct.

4. You did good work on the three.
I did #4 for you. Those are combinations.
${3 \choose 1}{5 \choose 2} \frac {4!} {2!} = (3)(10)(12)$,