1. ## Permutation

A cafe lets you order a deli sandwich your way. There are 6 choices of bread, 4 choices for meat, 4 choices for cheese, and 12 different garnishes.

How many different sandwich possibilities are there if you choose:
One bread, one meat, one cheese, and from 0 to 12 garnishes?

2. Tell us what have you tried? And where you have trouble?

3. Ive tried calculating the number of sandwiches from the bread, meat, and cheese, which is 6 x 4 x 4. But im confused of how to calculate the number of sandwiches with the garnishes.

4. For the garnishes, you have 13 choices, since you can choose
not to have one.
It's very unlikely you'd want to mix garnishes together.

5. Originally Posted by bori02082005
Ive tried calculating the number of sandwiches from the bread, meat, and cheese, which is 6 x 4 x 4. But im confused of how to calculate the number of sandwiches with the garnishes.
You can select the garnishes in $\displaystyle 2^{12}$ ways.
That is the number of subsets of a set of twelve items.

6. Thanks Plato,
a nice alternative to $\displaystyle \binom{12}{0}+\binom{12}{1}+\binom{12}{2}+......+\ binom{12}{12}$

1 garnish 'A' allows 2 choices 0 or A.

2 garnishes 'A' and 'B' allows 2 new additional choices B and (A,B), twice as many choices.

3 garnishes 'A', 'B' and 'C' allows 4 new additional choices C, (C,A), (C,B), (C,A,B), twice as many choices.

4 garnishes 'A', 'B', 'C', 'D'.... twice as many choices.

in general, number of ways of selecting groups of 0 to n $\displaystyle =2^n$

I can't begin to imagine a sandwich with 12 garnishes due to a
sensitive stomach..

7. Originally Posted by Archie Meade
I can't begin to imagine a sandwich with 12 garnishes due to a sensitive stomach..
Reality has absolutely nothing to do with mathematics.