# Combination Probability

1. Oct 3, 2007

### Kites

Here is the problem:

You have a sandwich shop. You can choose one of 6 different breads, 1 of four different cheeses, one of four different meats, and you can choose up to 12 garnishes, out of 0 to 12 garnishes.

Here is my solution so far.

In part A I calculated the amount of possible sandwiches without garnishes to be 96.

6*4^2 = 96

In part B i must calculated the following:

One bread, one meat, one cheese, and from 0 to 12 garnishes? (Remember there are 12 different choices for the garnish but you cannot choose the same garnish twice, so for each one of those 12 there are different possibilities.)

My first idea is to use the combination formula

nCr = n!/( r! (n-r)!)

I hesitate because I am confused of how to use it in respect to "from 0 to 12 garnishes." with there also being 12 garnish choices.

Thank you very much for clearing up some confusion.

2. Oct 3, 2007

### Kites

Solved. Thanks for the help anyway. :) 2^12

3. Oct 3, 2007

### Hurkyl

Staff Emeritus
You could use combinations. The number of ways to pick garnishes is

$$\sum_{r = 0}^{12} \binom{12}{r}$$

which simplifies to

$$\sum_{r = 0}^{12} \binom{12}{r} = \sum_{r = 0}^{12} \binom{12}{r} (1)^r (1)^{12 - r} = (1 + 1)^{12} = 2^{12}$$