If three dice are tossed, their sum could be 7. What is the probability of this?

Question

I'm guessing it's something to do with combinations?? I wrote down all the combinations of it equaling 7:
3, 3, 1 // 4, 2, 1 // 1, 5, 1
So I thought that since there are five possible numbers the three dice could be, it would be 5choose1 times 4choose1 times 3choose1, but I don't think that's right...

Answer 1

You left out the case 2,2,3

I would go with permutations, the total number of ways for the 3 dice to fall
= 6*6*6 = 216
To get a 7:
permute 223 in 3!/2! or 3 ways
permute 331 in 3 ways
permute 115 in 3 ways
permute 124 in 6 ways, for a total of 15 ways to get a 7

so prob(7) = 15/216
= 5/72