Stoby's opened on July 21,1980 with only one item on the menu, the stoby sandwich. If you are broke and decide to open a restaurant with only one item on the menu then a sandwich with 33,000. combinations would be a good choice, especially if it turns out to be popular . After 45 years the Stoby Sandwich is still the number one selling item on Stoby's menu.
How many sandwich combinations are possible on a Stoby Sandwich ?
Here's the math:
Calculate the possible sandwich combinations if you can choose one item from each of the four categories:
- 1 bread from 8 options
- 1 meat from 5 options
- 1 cheese from 5 options
- 1 topping from 3 options
Often you will see the answer, without any reference to the combinations equation C(n,r), as the multiplication of the number possible options in each of the categories. In this case we calculate:
8 × 5 × 5 × 3 = 600
possible sandwich combinations
In terms of the combinations equation below, the number of possible options for each category is equal to the number of possible combinations for each category since we are only making 1 selection; for example C(8,1) = 8, C(5,1) = 5 and C(3,1) = 3 using the following equation:
C(n,r) = n! / ( r!(n - r)! )
We can use this combinations equation to calculate a more complex sandwich problem.
Sandwich Combinations Problem with Multiple Choices
Calculate the possible combinations if you can choose several items from each of the four categories:
- 1 bread from 8 options
- 3 meats from 5 options
- 2 cheeses from 5 options
- 0 to 3 toppings from 3 options
Applying the combinations equation, where order does not matter and replacements are not allowed, we calculate the number of possible combinations in each of the categories. You can use the calculator above to prove that each of these is true.
- 1 bread from 8 options is C(8,1) = 8
- 3 meats from 5 options C(5,3) = 10
- 2 cheeses from 5 options C(5,2) = 10
0 to 3 toppings from 3 options; we must calculate each possible number of choices from 0 to 3 and get C(3,0) + C(3,1) + C(3,2) + C(3,3) = 8
Multiplying the possible combinations for each category we calculate:
8 × 10 × 10 × 8 = 6,400
possible sandwich combinations
How many possible combinations are there if your customers are allowed to choose options like the following that still stay within the limits of the total number of portions allowed:
- 2 portions of one meat and 1 portion of another?
- 3 portions of one meat only?
- 2 portions of one cheese only?
In the previous calculation, replacements were not allowed; customers had to choose 3 different meats and 2 different cheeses. Now replacements are allowed, customers can choose any item more than once when they select their portions. For meats and cheeses this is now a combinations replacement or multichoose problem using the combinations with replacements equation:
CR(n,r) = C(n+r-1, r) = (n+r-1)! / (r! (n - 1)!)
For meats, where the number of objects n = 5 and the number of choices r = 3, we can calculate either combinations replacement CR(5,3) = 35 or substitute terms and calculate combinations C(n+r-1, r) = C(5+3-1, 3) = C(7, 3) = 35.
Calculating cheese choices in the same way, we now have the total number of possible options for each category at
- bread is 8
- meat is 35
- cheese is 15
- toppings is 8
and finally we multiply to find the total
8 × 35 × 15 × 8 = 33,600
possible sandwich combinations!
Sandwich Combinations Problem
This is a classic math problem and asks something like 'How many sandwich combinations are possible?' and this is how it generally goes.
Calculate the possible sandwich combinations if you can choose one item from each of the four categories:
- 1 bread from 8 options
- 1 meat from 5 options
- 1 cheese from 5 options
- 1 topping from 3 options
Often you will see the answer, without any reference to the combinations equation C(n,r), as the multiplication of the number possible options in each of the categories. In this case we calculate:
8 × 5 × 5 × 3 = 600
possible sandwich combinations
In terms of the combinations equation below, the number of possible options for each category is equal to the number of possible combinations for each category since we are only making 1 selection; for example C(8,1) = 8, C(5,1) = 5 and C(3,1) = 3 using the following equation:
C(n,r) = n! / ( r!(n - r)! )
We can use this combinations equation to calculate a more complex sandwich problem.
Sandwich Combinations Problem with Multiple Choices
Calculate the possible combinations if you can choose several items from each of the four categories:
- 1 bread from 8 options
- 3 meats from 5 options
- 2 cheeses from 5 options
- 0 to 3 toppings from 3 options
Applying the combinations equation, where order does not matter and replacements are not allowed, we calculate the number of possible combinations in each of the categories. You can use the calculator above to prove that each of these is true.
- 1 bread from 8 options is C(8,1) = 8
- 3 meats from 5 options C(5,3) = 10
- 2 cheeses from 5 options C(5,2) = 10
0 to 3 toppings from 3 options; we must calculate each possible number of choices from 0 to 3 and get C(3,0) + C(3,1) + C(3,2) + C(3,3) = 8
Multiplying the possible combinations for each category we calculate:
8 × 10 × 10 × 8 = 6,400
possible sandwich combinations
How many possible combinations are there if your customers are allowed to choose options like the following that still stay within the limits of the total number of portions allowed:
- 2 portions of one meat and 1 portion of another?
- 3 portions of one meat only?
- 2 portions of one cheese only?
In the previous calculation, replacements were not allowed; customers had to choose 3 different meats and 2 different cheeses. Now replacements are allowed, customers can choose any item more than once when they select their portions. For meats and cheeses this is now a combinations replacement or multichoose problem using the combinations with replacements equation:
CR(n,r) = C(n+r-1, r) = (n+r-1)! / (r! (n - 1)!)
For meats, where the number of objects n = 5 and the number of choices r = 3, we can calculate either combinations replacement CR(5,3) = 35 or substitute terms and calculate combinations C(n+r-1, r) = C(5+3-1, 3) = C(7, 3) = 35.
Calculating cheese choices in the same way, we now have the total number of possible options for each category at
- bread is 8
- meat is 35
- cheese is 15
- toppings is 8
and finally we multiply to find the total
8 × 35 × 15 × 8 = 33,600
possible sandwich combinations!
References
[1] Zwillinger, Daniel (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. 206, 2003.
For more information on combinations and binomial coefficients please see Wolfram MathWorld: Combination.