# ice cream flavors, cones, and toppings. A counting question.

There are 5 distinct ice cream flavors, 5 distinct cones, and 5 distinct toppings. I gather all 15 distinct supplies. There are 5 people. Each person has to get a cone, flavor, and topping, and I can give them any combination of cone, flavor, topping that I want. I don't have to complete one ice cream at a time. For example, I can give everyone a cone, and then the flavor, and then the topping.

However, I have to give a cone first before I give the ice cream, and I have to give ice cream before I give a topping.

How many ways can I hand out supplies under these constraints? (I can only do one action at a time).

My thoughts: I have to give a cone first, and there are 5 distinct cones, so 5 choose 1. Then, I can give a cone, or the flavor as the second step, so another 5 options. But I'm stuck on what to do from here.

Call the people $$A,B,C,D,E$$. If you make a string of fifteen characters with three of each letter, each string is one way to hand out the supplies. The first time you see $$A$$ you give $$A$$ a cone. The second time you see $$A$$ you give $$A$$ ice cream. The third time you see $$A$$ you give $$A$$ a topping. To count the strings, you choose $$3$$ positions for the $$A$$s, then three of the remaining $$12$$ for $$B$$s, and so on. This gives $${15 \choose 3}{12 \choose 3}{9 \choose 3}{6 \choose 3}{3 \choose 3}=\frac {15!}{3!^5}=168\ 168\ 000$$ orders you can hand out the supplies. Presuming you can only give one person each kind of ice cream, etc. there are $$5!$$ ways to assign the cones to people, $$5!$$ ways to assign the ice creams, and $$5!$$ ways to assign the toppings, so the grand total ways you can hand out stuff is $$\frac {15!}{3!^5}5!^3=290\ 594\ 304\ 000\ 000$$