8. How many different pizzas can be made with 4 types of crust, 2 types of cheese, and 7 different toppings if each pizza has 1 crust, 1 cheese, and 1 topping?

Problem statement:  The menu at a pizza place lets you pick one of two crust types, one of three sauces, one of five types of cheese, and any number of seven different toppings. How many possible different pizzas can you order that have exactly three different toppings?

Be very careful to read this question critically and to enumerate the number of items correctly:

 "one of two crust types"      2 possibilities

 "one of three sauces"   3 possibilities

 "one of five types of cheese"  5 possibilities

 "any number of seven different toppings"    

      many possibilities (none, any combination of {a,b,c,d,e,f or g})

However, "exactly three different toppings" (not cheeses or sauces) means that you are choosing 3 of 7 toppings (and that's all that this problem asks you).

So, nCr = n! / ((n-r)!r!)        ["n choose r";  n=7; r=3]

    7C3 = 7! / (4!3!)

    7C3 = 5040/144

    7C3 = 35

it’s supposed to be -4+45=41

step-by-step explanation:

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204