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Hitman42 [59]
3 years ago
15

At LaMar’s birthday party, his parents served slices of plain, sausage, or pepperoni pizza. They also served cups of milk, orang

e 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?
Mathematics
2 answers:
snow_lady [41]3 years ago
8 0

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.


ziro4ka [17]3 years ago
6 0
I GOT SOME UPDATE SUPERBLACK IS DEAD
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Therefore, the number of possible lineups by Team A is

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