Question

11. A pizza parlor has four types of crust, 12 different toppings, and three different cheese blends to choose from. In how many ways could you order a three-topping pizza (one type of crust, three different toppings, and one cheese blend)

Answer 1

Answer: 1320

Step-by-step explanation:

Given: In pizza parlor, Number of types of crust = 4

Number of types of different toppings = 12

Number of types of cheese blends = 3

Now, the number of ways to choose a pizza with one type of crust, three different toppings, and one cheese blend

= $^{4}C_1\times\ ^{12}C_3\times\ ^3C_1$  

= $4\times \ \dfrac{12!}{3!(12-3)!}\times 3\ \ \ \ [^nC_r=\dfrac{n!}{r!(n-r)!},\ ^nC_1=n]$

= $12\times\dfrac{12\times11\times10\times9!}{2\times6\times9!}$

= $12\times11\times10= 1320$

Hence, the required number of ways = 1320

Answer 2

Using the Fundamental Counting Theorem, it is found that you could order a three-topping pizza in 144 ways.

What is the Fundamental Counting Theorem?

It is a theorem that states that if there are n things, each with $n_1, n_2, \cdots, n_n$ ways to be done, each thing independent of the other, the number of ways they can be done is:

$N = n_1 \times n_2 \times \cdots \times n_n$

In this problem, we have that a pizza parlor has four types of crust, 12 different toppings, and three different cheese blends to choose from, hence:

$n_1 = 4, n_2 = 12, n_3 = 3$

Then:

$N = 4 \times 12 \times 3 = 144$

You could order a three-topping pizza in 144 ways.

More can be learned about the Fundamental Counting Theorem at https://brainly.com/question/24314866