Math Question Combinations

Question

A restaurant has 3 appetizers, 8 Main courses, 4 sides, 4 vegetable choices, and 6 dessert choices. A diner enters the restaurant and orders and eats one item at a time. The diner orders one item from three different categories. In how many ways could the diner have placed the order based on the category?

Answer 1

This problem is an example where we will need to count different groups - since the different categories all have their own number of items, and we only choose some of the categories to eat from, so we will need to first find all the possible 3 category combinations, then for each 3 category combination, we count how many different three item orders exist.

First lets find all 3 category combinations, I'll refer to the category as its first initial:

AMS, AMV, AMD, ASV, ASD, AVD, MSV, MSD, MVD, SVD.

Now for each of these 3 category combinations, we can find the number of possible orders by multiplying the number of items in each category.

AMS: 3*8*4; AMV 3*8*4; AMD: 3*8*6; ASV:3*4*4; ASD: 3*4*6; AVD: 3*4*6; MSV: 8*4*4; MSD: 8*4*6;

MVD: 8*4*6; SVD: 4*4*6.

Then we add these products together to get the total number of possible orders. We add these counts together because each count belongs to its own unique group of categories, whereas multiplying counts is done when we are counting combinations in a particular group. This is an important distinction in combinatorics.

We are left with the final answer:

(3*8*4) + (3*8*4) + (3*8*6) + (3*4*4) + (3*4*6) + (3*4*6) + (8*4*4) + (8*4*6) + (8*4*6) + (4*4*6) = 1136.

So there are 1136 possible orders assuming the order is one item each from 3 different categories.

(I would check my calculations to be careful - I looked over it a few times but there's a lot of numbers to juggle here so making a mistake is very easy.)