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PS: It's not \[\binom{12}{3}\]

220

Nope.

1320?

i'm quite sure it's 220 tho

i was never good at combinatorics =/ 91?

I believe that is the answer. Good job. Mind explaining how you got that to everyone?

what's the answer?

That's probably the best way to look at it.

Bingo! This is also called the "stars and bars" problem if anyone wants to look it up.

typo, twelve 1's and two +'s

http://spot.colorado.edu/~kearnes/F09/distrib_sol.pdf

@robtobey did you take that class?

Well, rather the class two years later.

No discrete at my school either ;;

Interesting. I would love to take a class based purely on problem solving.

there's a formula for identical problems: n+r-1Cr-1

guess i'm too late

Nice try @ninhi5 :)

nice question, havent taken combination and permutation for ages

There is a generating function approach to solve this one too.

I have already shown that

Sorry, I didn't read the whole thread.