Math problem: how many different two scoop ice cream cones are possible?
"An icecream shop offers three different cones, ten different flavors, and four different toppings. How many different two scoop cones with up to two distinct toppings are possible, if order does not matter?"
Evidently the answer is 2295. The answer key to the test says there are (10C2)*3 combinations with no toppings, (10C2)*3*4 with one, and (10C2)*3*12 with two, which add to 2295.
But I'm confused. 10C2 doesn't allow the two scoops to be the same flavor, and why are there 12 options for two toppings instead of 6 if order isn't supposed to matter?
- MercyLv 74 weeks ago
- PuzzlingLv 74 weeks ago
I agree. The question isn't very well stated and the answer is confusing.
It seems to assume that you can't order two scoops of the same flavor. So that explains the 10C2. There would be 10 additional ways to pick a double-scoop of a single flavor, if the same flavor were allowed. So let's assume you can't pick a double-scoop of the same flavor.
As for the toppings, I agree there as well. You can choose no toppings (1 way), or 1 topping (4C1 = 4 ways) or 2 toppings (4C2 = 6 ways).
I would say the answer is:
3C1 = 3 choices of cone
10C2 = 45 choices of two different flavors of ice cream (no duplicates)
4C0 + 4C1 + 4C2 = 1 + 4 + 6 = 11 ways to pick toppings (no duplicates)
3 * 45 * 11 = 1485
I don't know why they would have 12 which is 4P2. That would assume the order of toppings does matter.
It's a bad question. And they seem to have a mistake in the answer.