# Figuring out # of possible combos of 3

1. Sep 7, 2004

### Math Is Hard

Staff Emeritus
A friend asked me this question, and I am not sure I know how to answer it:

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I have 15 items, and I want to figure out how many unordered combinations of three there are. Any ideas?

a b c d e f g h i j k l m n o

Example:
abc bcd cde
abd bce cdf
abe bcf cde
abf bcg cdg

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The way I think it can be done is using this formula:
$${n\choose k}={n!\over k!(n-k)!}$$

so you would have $${15!\over 3!12!}$$ or 455 possible combinations.
But I am not really confident in this because I never studied this formula in a class - and I don't know if I am using it correctly.

Thanks!
-MIH

2. Sep 7, 2004

### HallsofIvy

Staff Emeritus
Think about how you would choose such combinations. Start by choosing the first letter: there are 15 ways to do that. Now choose the second letter: 14 ways. Finally, choose the third letter: 13 ways. There are 15*14*13 ways. You could write that as 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1/12*11*10*9*8*7*6*5*4*3*2*1= 15!/12!= 15!/(15-3)!

But that counts "abc" and "bac" as separate which you don't want to do. For each 3 letters there are 3! ways to get those same letters in different orders. To account for that, divide the answer above by 3!. That gives $${15!\over 3!12!}$$
exactly as you suggest. The same analysis for k objects chosen from n gives $${n\choose k}$$ (which, in tex form is written "n choose k"!).

3. Sep 7, 2004

### Math Is Hard

Staff Emeritus
Thanks so much, Halls!!!
I appreciate you taking the time to show me how that works.

4. Sep 14, 2004

### Sharayah

Hey does anyone know physics???

5. Sep 14, 2004

### Gokul43201

Staff Emeritus
Sharayah, if you have a physics question, post in under General Physics, and create a New Thread.