How many sandwich combinations can you create with different garnish options?

Click For Summary
SUMMARY

The discussion focuses on calculating the total number of sandwich combinations possible at a sandwich shop, given specific choices for bread, cheese, meat, and garnishes. The calculation for sandwiches without garnishes results in 96 combinations, derived from the formula 6 * 4^2. For garnishes, the total combinations from 0 to 12 garnishes are calculated using the binomial theorem, resulting in 2^12, which accounts for all possible selections of garnishes from the 12 available options.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically the combination formula nCr.
  • Familiarity with the binomial theorem and its application in counting problems.
  • Basic knowledge of multiplication principles in combinatorics.
  • Ability to perform factorial calculations.
NEXT STEPS
  • Study the binomial theorem in detail to understand its applications in combinatorial problems.
  • Learn about advanced combinatorial techniques, such as permutations and combinations.
  • Explore practical applications of combinatorial mathematics in real-world scenarios.
  • Practice solving similar problems involving combinations and permutations with different constraints.
USEFUL FOR

Mathematicians, students studying combinatorics, and anyone interested in optimizing choices in scenarios involving multiple options and constraints.

Kites
Messages
39
Reaction score
0
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.
 
Physics news on Phys.org
Solved. Thanks for the help anyway. :) 2^12
 
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}<br /> = \sum_{r = 0}^{12} \binom{12}{r} (1)^r (1)^{12 - r}<br /> = (1 + 1)^{12}<br /> = 2^{12}<br />
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K