# Homework Help: Counting problem involving anagrams

1. Nov 7, 2012

### V0ODO0CH1LD

1. The problem statement, all variables and given/known data

How many anagrams with 4 distinct letters and that have two of the letters "a", "b" and "c" can you make using the first ten letters of the alphabet?

2. Relevant equations

3. The attempt at a solution

First I assume that by anagram they mean letters arranged in any order (like as if any two lettered word had two anagrams and any three lettered word had six).

Now as far as solving the problem goes, I don't even know where to start.. I though of using factorials for the 4 distinct letters, but I'm not even sure of that.

2. Nov 7, 2012

### SammyS

Staff Emeritus
An anagram of four letters is simply a four character string of letters of the alphabet. It doesn't necessarily spell anything.

3. Nov 7, 2012

### LCKurtz

Think about how many ways you can pick two letters from a,b,c and how many ways to pick the other two and how many ways to arrange them.

4. Nov 7, 2012

### V0ODO0CH1LD

I thought about starting from simpler scenarios and work from there. Like; if I just wanted all the anagrams that can be formed using the first ten letters of the alphabet it would be 10^10; right? But then I want to subtract all the anagrams that have 7, 8, 9 or 10 repeated letters.

For the 10 repeated letters it's easy, just one anagram for each letter that's 10.
aaaaaaaaaa
bbbbbbbbbb
...
jjjjjjjjjjjjjjjjjjj

For 9 it's 10 for each letter, but for each 10 of those there is 9 options for the 10th unrepeated letter. So..

aaaaaaaaab
aaaaaaaaba
...
baaaaaaaaa
...
aaaaaaaaaj
aaaaaaaaja
...
jaaaaaaaaa

So far I have 10 + 900. But I think I am overcomplicating things. There must be an easier way to do this..

5. Nov 7, 2012

### SammyS

Staff Emeritus
The problem states that the anagrams have 4 letters, and the letters are distinct. -- No repeats and only 4 letters in length.

6. Nov 7, 2012

### V0ODO0CH1LD

Ah! Damn my interpretation skills.

Okay; so I have three ways to select two letters out three. Then I have 12 ways to organize those in a four character string (taking into consideration order matters). That gives me 36 ways to get two of the letters a, b and c in my anagram; right? But I still have 2 spots left and for the first I could have any of the eight remaining letters minus the one left out from a, b and c. And finally for the last spot I have six options.

So that is (3*12)*7*6 = 1512. Is that correct or did I forget something?

7. Nov 7, 2012

### LCKurtz

That is correct. But it is more direct expressed like this:$$\binom 3 2 \binom 7 2 4!$$where the first binomial is the number of ways of choosing the 2 from the 3, the next binomial is the number of ways of choosing the other 2 from the remaing 7, and the factorial is the number of ways to arrange the 4 after you have them.

8. Nov 7, 2012

### V0ODO0CH1LD

Okay! Thanks!