# Combinations! How many.

1. Jun 2, 2007

### Elruso

Combinations! How many.........

1. The problem statement, all variables and given/known data
How many letter combinations with 9 letters are you able to make with following letters : M-A-T-E-M-A-T-I-K?

2. Relevant equations
Well its pretty obvious you need to use Combinations.

Please explain how you solve this problem, don't write use combinations .
I need to know how you think and from which angle you "attack" the problem.

2. Jun 2, 2007

### Dick

You have 9 letter positions to fill. First lets place the M's. There are two of them, so I have C(9,2) ways. Now lets do the A's. There 2 of them and 7 places left to fill, so I have C(7,2) ways. So far I've got C(9,2)*C(7,2). Can you finish?

3. Jun 3, 2007

### Elruso

So iit´s C(9,2)*C(7,2)*C(5,2)*3*2*1?

In my math book the answer is C(9,2)*C(7,2)*5!..... which i find a little strange.

4. Jun 3, 2007

### danago

When i did it, i got the same answer as you, and then to check, i got mathematica to output every single permutation of those letters into a list. That list contained 45360 elements, so unless ive misunderstood the question, it seems that you may be right.

5. Jun 3, 2007

### Hurkyl

Staff Emeritus
Did you notice those are the same thing?

Incidentally, it seemed most clear to me to write the answer as
9! / (2! * 2! * 2!),​
or, as a multinomial coefficient,
$$\binom{9}{2 \ 2 \ 2 \ 1 \ 1 \ 1} .$$​

Last edited: Jun 3, 2007
6. Jun 3, 2007

### Dick

They aren't the same thing. They differ by a factor of two. Whoever wrote the solution seems to have miscounted the number of doubled letters.

7. Jun 3, 2007

### Hurkyl

Staff Emeritus
Ah, right. This is what was written:
C(9,2)*C(7,2)*C(5,2)*3*2*1​
and this is what I thought I read:
C(9,2)*C(7,2)*C(5,2)*3!*2!*1!​