how would I get this answer?

Question

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?

Answer 1

You have different combinations

for bread you have 3 choices

For meat, you have 4 choices one of the three meats, or no meat.

For the cheeses, you have (4 choose 2) = 4!/((4-2)2!) = 4x3/2=6 possible combinations of 2 cheeses, or you can choose only 1 cheese = 4 choices, or you can choose no cheese (1 choice), so for cheese, you have 6+4+1 = 11 choices

For toppings, you have (8 choose 3) = 8x7x6/(3x2x1) = 56 combinations of three toppings, or you can choose 2 = 8x7/2 = 28 choices, or you can choose 1 = 8 choices or you can choose none = 1 choice. Total choices is 56+28+8+1 = 93 choices.

Overall you have 3x4x11x93 = 12276 possible sandwich combinations.

Answer 2

OPtions:
B=3
M=4 (includinf none)
C=5 (including none)
T=9 (including none)

Bread options=3
Meat options=4
Cheese options= 5*4/2*1=10

Topping options=9*8*7/(3*2*1)=84

Total options=3*4*10*84=10080

Answer 3

There are 3 ways to choose bread, AND
4 ways to choose meat (including no meat), AND
(6 ways to choose 2 cheeses, OR
4 ways to choose 1 cheese, OR
1 way to choose no cheese), AND
(56 ways to choose 3 toppings, OR
28 ways to choose 2 toppings, OR
8 ways to choose 1 topping, OR
1 way to choose no toppings).

In summary, there are 3 ways to choose the bread, 4 ways to choose the meat, 11 ways to choose the cheese, and 93 ways to choose the toppings. Multiply them all together and you get 3*4*11*93=12,276 possible sandwiches.

Note: I used the binomial coefficients

C(n,k) = n!/[k!(n-k)!],

which tells you the number of ways to choose k objects from a set of n.

Answer 4

what he said...