Calculate Combinations of 10 Items (Max 3) - Paul

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To calculate the number of sandwich combinations with 10 items, where one bread, one meat, up to two cheeses, and up to three toppings can be selected, the fundamental principle of counting is applied. The total combinations for each category are calculated separately, with the bread and meat selections being straightforward, while the cheese and topping selections require combination formulas. It is emphasized that order does not matter for toppings and cheeses, making combinations the appropriate method. The discussion suggests breaking the problem into cases for clarity and accuracy in calculations. Ultimately, the total number of unique sandwich combinations can be determined by summing the results from each case.
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Say i needed to calculate the different number of combinations there are if you have 10 items and can pick up to 3 of them. e.g you buy a sandwhich and have Ketchup, Mustard, Relish, Lettuce, Pickles, Sour Cream, Cream Cheese, Olives as available toppings but you can only choose up to three. how many combinations are there?

Thanks
Paul
 
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What have you done so far to solve your problem?
 
Well the entire problem is as follows:
You are at a corner deli with a craving for a sandwich, here's the menu:
Breads: wheat, rye, white
Meats: turkey, ham, salami
Cheeses: American, Swiss, Cheddar, Gouda
Toppings: Ketchup, Mustard, Relish, Lettuce, Pickles, Sour Cream, Cream Cheese, Olives

You can only get one kind of bread (you have to have bread, no low-carb diet).
Per sandwich you're only allowed up to one kind of meat, up to two kinds of cheese, and up to three toppings. This means you can have none of the above options. The minimum required food is bread with nothing on it.

How many different options do you have?



So far i have worked out that there are 120 combinations up to the cheese selection (unless i have made an error)
 
Use combination and fundamental principle of counting
 
think of how many choices you have for the each selection. Then you can use the rule of product or the formula P(n,r)=n!/(n-r)! might help you out.
 
buzzmath said:
think of how many choices you have for the each selection. Then you can use the rule of product or the formula P(n,r)=n!/(n-r)! might help you out.

I don't think this would be a situation to use a permutation, a combination would be better, i think the problem would consider a sandwich with lettuce and chese to be the same as a sandwich with cheese and lettuce, so order doesn't matter. Don't read too much into that example because it doesn't really fit what the problem is asking but it gets my point across..
 
Solve in three cases. In each case you can use combination and further you can add the three cases to get the result.
 
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