# A restaurant offers 5 choices of appetizer, 10 choices of main meal and 4 choices of dessert.

A customer can choose to eat just one course, or two different courses, or all three courses. Assuming all choices are available, how many different possible meals does the restaurant offer?

So here is what I thought of. Since the question said that a customer can choose one, two or all of three courses, I would use a Venn diagram for this matter.

First of all I created a situation where A would represent appetizer, B would be main meal and C would be desserts. So if I use a Venn diagram, the intersection in this case which can represent the customer choosing from one up to all three courses. The part where I got confused is will the intersection point between each 3 section of the Venn diagram have different unknowns? Help me out.

• Put all of the information in the post, don't have the question starting in the title. Also, hint: Find the total number of combinations for three courses, total number for two courses and the total number for one course (clearly, $1$ course is $5+10+4=19$) and add the totals together. Venn diagrams visually describe the relationships between sets, it might be a more difficult approach in this case. – Mattos Sep 10 '15 at 14:55
• Here probability tag is unnecessary. Where does the concept of probability arises? – user249332 Sep 10 '15 at 15:01
• @SubhadeepDey So is the permutations tag since order of courses is irrelevant here. – Rohcana Sep 10 '15 at 15:02
• @Anachor, Yes, of course, only Combinations and Combinatorics could suffice. – user249332 Sep 10 '15 at 15:08

## 2 Answers

The problem has nothing to do with Venn diagrams. You are calculating the intersections of appetizer, main and dessert, which means you are calculating the number of items which can serve as all three type of courses, which has nothing to do with the problem.

I have given a solution below, but if you want have another go at it, Here's a hint. I am only restating the problem.

You have $5$ types of item A, $10$ types of item B, $4$ types of item C, You have to count the number of nonempty sets which contains at most one item of each type.

A possible solution:

For each type of course we calculate the possibilities for the customer. For the appetizer, he can choose any one of five or choose none at all. Therefore he has $6$ choices for the appetizer. Similarly he has $11$ choices for main course, and $5$ for the dessert for a total of $$6 \times 11 \times 5 = 330$$.

However, there is one case, where we are choosing none of the courses, we must exclude it (we don't want to keep the customer hungry, do we?) , hence the answer is $329$

Ways of choosing 3 course meal 5*10*4=200 Ways of choosing 2 course meal (5*10)+(5*4)+(10*4)=110 Ways of choosing 1 course meal 5+10+4=19

Total possible ways of choosing at least 1 option: =(ways of choosing 3 course meal)+(ways of choosing 2 course meal)+(ways of choosing 1 course meal)= 200+110+19=329

## protected by Community♦Nov 2 '18 at 1:30

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?