Question

A pizza place offers 4 different cheeses and 10 different toppings. In how many ways can a pizza be made with 1 cheese and 3 toppings?

Answer 1

480
360
240
120

Answer 2

wow ,not bad -- so many

Answer 3

haha, yes, but whats the answer haha

Answer 4

Just a guess, is it:\[\left(\begin{matrix}4 \\ 1\end{matrix}\right)\left(\begin{matrix}10 \\ 3\end{matrix}\right)\]

Answer 5

yeah, i have no clue.

Answer 6

So that would be the first option.

Answer 7

I just tried something - you can arrange three components in 6 different ways
1, 2, 3
1, 3, 2
2, 1, 3
2, 3, 1
3, 1, 2
3, 2, 1
You would also get this by calculating 3!, which is 3*2*1 = 6

Answer 8

@seraphic_topaz can you write out what one gets from (10 over 3) in concrete formula?

Answer 9

we have 10 choices for topping, we do this three times, you're not supposed to use the same topping three times, I'm sure.

so first you could pick 10 different toppings, the next time you could pick 9 different toppings, and then finally there is 8 to choose from left.

It doesn't matter IN WHICH ORDER you picked them, this is crucial for probabilities.

Answer 10

so there's 10*9*8 different possibilities how the putting of toppings could take place. That's 720. However one combo can also be put together six different ways. so is the 720 all the different ways or just one?

Answer 11

if you have 10 SEPERATE paths, and then in every one of them you get 9 NEW PATHS, I get the feeling that it DOES count EVERY DIFFERENT PROCEEDING.

tomato + onion + salad
would not be the same choice as
onion + salad + tomato

The answer is (720/6) * 4 -> 120*4= __480__.

Answer 12

The formula of \[\left(\begin{matrix}10 \\ 3\end{matrix}\right)\] is 10!/(3!7!)