A pizzeria sells small, medium, and large pizzas, and...

Hey swerve,

Let's break this problem down into the different options the customer has.

Size
All three pizzas the customer orders must be the same size: small, medium, or large. So the customer has 3 options for size.

Toppings
This one is a little more complicated.

One option is for all three pizzas to have the same topping. There are 5 different toppings, so the customer has 5 different options for toppings if all three pizzas have the same topping.

The other options is for all three pizzas to have different toppings. So we are looking for all of the different combinations of 3 toppings we can get from the 5 available toppings — in other words, we have a 5 choose 3 combination:
$$\frac{5!}{\left(5-3\right)!3!}=\frac{5!}{2!3!}=\frac{5\cdot4\cdot3!}{2\cdot1\cdot3!}=\frac{5\cdot4}{2\cdot1}=\frac{20}{2}=10$$
So the customer has 10 different options for toppings if all three pizzas have different toppings.

If the customer has 5 options for all the same toppings and 10 options for all different toppings, they have 5+10=15 options for toppings in total.


Bringing this information together, if the customer has 15 options for toppings for each of their 3 options for size, the customer has 3*15=45 unique options for pizza orders, or answer choice C.

swerve wrote:A pizzeria sells, small, medium, and large pizzas, and offers five different toppings. Its "Three for Thursday" delivery special allows diner to order three one-topping pizzas for the price of one, but with one restriction: all three pizzas must be the same size, and the topping for each pizzas must either be the same for all three pizzas or different for each pizza. How many unique orders are available with this special?

A. 25
B. 35
C. 45
D. 55
E. 75

Since the 3 pizzas in the special must be the same size, we can determine the number of ways the special consists of all small pizzas first and then multiply that result by 3 (since the pizzas could be also medium or large besides small).

If the 3 small pizzas have the same topping, then there are 5 ways to order them since there are 5 different toppings. If the 3 small pizzas have different toppings, then there are 5C3 = (5 x 4 x 3)/(3 x 2) = 10 ways to order them.

Therefore, there are 5 + 10 = 15 ways to order the 3 small pizzas and 15 x 3 = 45 ways to order 3 same-sized pizzas.

Answer: C