Question

At LaMar’s birthday party, his parents served slices of plain, sausage, or pepperoni pizza. They also served cups of milk, orange juice, apple juice, or lemonade. If each guest at the party chose one slice of pizza and one type of drink, how many different combinations were possible?

Answer 1

The choices on pizza are plain, sausage, or pepperoni. Thus, there are three types of pizza and a guest can choose only one type.

Likewise, the choices on drinks are milk, orange juice, apple juice, or lemonade. Thus, there are four types of drinks and a guest can choose only one type.

This is a typical case of Combination where for pizza the guest has to choose 1 from 3 and for drinks the guest has to choose 1 from 4. Thus, we have the following setup:

For Pizza: $^nC_r=^3C_1$

For Drinks: $^nC_r=^4C_1$

Thus, the total number of combinations possible will be:

$^3C_1 \times ^4C_1=\frac{3!}{2!1!}\times \frac{4!}{3!1!}$

$=\frac{3\times 2!}{2!}\times \frac{4\times 3!}{3!}=3\times 4=12$

Therefore, 12 is the answer.

Answer 2

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