- #1

goraemon

- 67

- 4

## Homework Statement

An ice cream shop has a special on banana splits, and Xing is taking advantage of it. He’s astounded at all the options he has in constructing his banana split:

· He must choose three different flavors of ice cream to place in the asymmetric bowl the banana split is served in. The shop has 20 flavors of ice cream available.

· Each scoop of ice cream must be topped by a sauce, chosen from six different options. Xing is free to put the same type of sauce on more than one scoop of ice cream.

· There are 10 sprinkled toppings available, and he must choose three of them to have sprinkled over the entire banana split.

a) How many different ways are there for Xing to construct a banana split at this ice cream shop?

b) Suppose that instead of requiring that Xing choose exactly three sprinkled toppings, he is allowed to choose between zero and three sprinkled toppings. In this scenario, how many different ways are there for him to construct a banana split?

## Homework Equations

P(n,k), C(n,k), product rule, sum rule?

## The Attempt at a Solution

(a) I'm pretty sure I got the answer for this one but would like to double-check. C(20, 3) * 6^3 * C(10, 3).

(b) The first two choices should be the same as in part (a) above: C(20, 3) * 6^3

For the final part where Xing must choose anywhere between 0 and 3 toppings (which I assume are distinct toppings), I figured:

1 way to choose zero toppings + 10 ways to choose 1 topping + C(10, 2) ways to choose 2 toppings + C(10, 3) ways to choose 3 toppings = 176 total ways to choose between 0 and 3 toppings

Therefore, final answer: C(20, 3) * 6^3 * 176

Is this right? I'm a bit rusty on combinatorics / probability and not sure whether / why I should use the sum rule rather than the product rule. Also, any other places where I messed up? Thanks.