# Combinatorics - counting problem

• goraemon
In summary: So in this case, since you can choose any number of toppings, you have to use the sum rule. The product rule would only apply if you had to choose exactly 3 toppings. In summary, there are C(20, 3) * 6^3 * (C(10,0)+C(10,1)+C(10,2)+C(10,3)) total ways for Xing to construct a banana split at this ice cream shop.
goraemon

## 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.

Looks right to me. Just for brevity of expression, you can write the number of ways of choosing toppings in (b) as

C(10,0)+C(10,1)+C(10,2)+C(10,3)

Sums are used when you have to choose one thing or the other
Products are used when you have to choose one of both
ie sum <--> OR, product <--> AND

## What is combinatorics?

Combinatorics is a branch of mathematics that deals with counting and arranging objects or elements in a systematic manner.

## What are counting problems in combinatorics?

Counting problems in combinatorics involve finding the number of possible outcomes or arrangements of a given set of objects or elements. These problems often involve using techniques such as permutations and combinations.

## What is the difference between permutations and combinations?

Permutations involve arranging a set of objects or elements in a specific order, whereas combinations involve selecting a subset of objects or elements without regard to order.

## How do I solve counting problems in combinatorics?

The most common approach to solving counting problems in combinatorics is by using the fundamental principle of counting, which states that the total number of outcomes for a sequence of events is equal to the product of the number of outcomes for each event. Other techniques such as permutations, combinations, and the binomial theorem can also be used depending on the problem.

## What are some real-world applications of combinatorics?

Combinatorics has many practical applications in fields such as computer science, statistics, economics, and biology. Some examples include analyzing the probability of genetic outcomes, designing efficient computer algorithms, and predicting the likelihood of success in a given situation.

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