Chapter 11.7, Problem 32E

a.

To determine

To find the number of pizzas having 1 meat topping and 2 different vegetable toppings can be selected if there are 4 meat toppings and 6 vegetable toppings

Answer to Problem 32E

60 Pizzas having 1 meat topping and 2 different vegetable toppings.

Explanation of Solution

Given information:

There are 4 meat toppings and 6 vegetable toppings

Calculation:

  $\begin{array}{l}{}_4{C}_1{\times }_6{C}_2\\ {}_n{C}_r{\times }_n{C}_r=\frac{n!}{r!\left(n-r\right)!}\times \frac{n!}{r!\left(n-r\right)!}\text{ [Use combinations formula]}\\ \text{ = }\frac{4!}{1!\times 3!}\times \frac{6!}{2!\times 4!}\\ \text{ =}\frac{4\times 3}{1\times 3}\times \frac{6\times 5\times 4\times 3\times 2}{2\times 4\times 3\times 2\times 1}\text{ [Divide out common]}\\ \text{ =4}\times \text{15}\\ \text{ =60 [Simplify}\text{.] }\end{array}$

b.

To determine

To find the number of pizzas having 2 different meat toppings and 1 vegetable topping can be selected if there are 4 meat toppings and 6 vegetable toppings

Explanation of Solution

36 pizzas having 2 different meat toppings and 1 vegetable topping.

Given information:

There are 4 meat toppings and 6 vegetable toppings

Calculation:

  $\begin{array}{l}{}_4{C}_2{\times }_6{C}_1\\ {}_n{C}_r{\times }_n{C}_r=\frac{n!}{r!\left(n-r\right)!}\times \frac{n!}{r!\left(n-r\right)!}\text{ [Use combinations formula]}\\ \text{ =}\frac{4!}{2!(4-2)!}\times \frac{6!}{1!(6-1)!}\\ \text{ =}\frac{4!}{2!\times 2!}\times \frac{6!}{1!\times 5!}\text{ [Divide out common]}\\ \text{ =6}\times \text{6}\\ \text{ =36 [Simplify}\text{.]}\end{array}$

c.

To determine

To find the number of pizzas having 3different toppings if there are 4 meat toppings and 6 vegetable toppings

Answer to Problem 32E

120 pizzas having 3 different toppings.

Explanation of Solution

Given information:

There are 4 meat toppings and 6 vegetable toppings

Calculation:

  $\begin{array}{l}{}_{10}{C}_3\\ {}_n{C}_r=\frac{n!}{r!\left(n-r\right)!}\text{ [Use combinations formula]}\\ {}_{10}{C}_3=\frac{10!}{3!\left(10-3\right)!}\\ {}_{10}{C}_3=\frac{10!}{3!\times 7!}=\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3}{3\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\text{ [Divide out common]}\\ {}_{10}{C}_3=120\text{ [Simplify}\text{.]}\end{array}$