# Permutation

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• Jan 17th 2010, 10:00 AM
bori02082005
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?
• Jan 17th 2010, 10:23 AM
Plato
Tell us what have you tried? And where you have trouble?
• Jan 17th 2010, 10:37 AM
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.
• Jan 17th 2010, 10:46 AM
Archie Meade
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.
• Jan 17th 2010, 11:47 AM
Plato
Quote:

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 $2^{12}$ ways.
That is the number of subsets of a set of twelve items.
• Jan 17th 2010, 03:55 PM
Archie Meade
Thanks Plato,
a nice alternative to $\binom{12}{0}+\binom{12}{1}+\binom{12}{2}+......+\ binom{12}{12}$

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 $=2^n$

I can't begin to imagine a sandwich with 12 garnishes due to a
sensitive stomach..
• Jan 17th 2010, 04:17 PM
Plato
Quote:

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.