Math question Snow cones?

If you only choose one syrup, clearly there are 12 choices.

If you choose two syrups, you have 12 choices for the first syrup and then 11 choices for the second syrup (because duplicate flavors are not allowed.) So that gives 12 × 11 choices.

But some choices amount to the same combination. If you choose chocolate then vanilla that's the same flavor combination as if you chose vanilla then chocolate! Each combination of 2 flavors can be chosen as "a then b" or "b then a". To compensate for this, we divide the number of choices by 2, giving 12 × 11 / 2 = 66 combinations.

If you choose three syrups, you have 12 choices for the first syrup, 11 choices for the second syrup, and 10 choices for the third. That gives 12 × 11 × 10 choices.

However, just like above, some combinations of flavors can be chosen more in more than one way. If we have three flavors -- a, b, c -- they can be chosen as:

a then b then c

a then c then b

b then a then c

b then c then a

c then a then b

c then b then a

That's 6 ways. (We could have calculated that without writing out all the choices by noting that there are 3 ways to make the first choice, then 2 ways to make the second choice, and just 1 way to make the third choice, giving 3 × 2 × 1 = 6.)

Thus, we divide the 12 × 11 × 10 choices by 3 × 2 × 1, which gives 220 combinations.

So in total, there are 12 + 66 + 220 = 298 flavor combinations.

The short answer is C(12,1) + C(12,2) + C(12,3), where C(n,m) = n! / (m! (n - m)!) is the number of combinations of m objects chosen from n distinct objects, where the order of choice does not matter.