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lys-0071 [83]
3 years ago
9

Jane must select three different items for each dinner she will serve. The items are to be chosen from among five different vege

tarian and four different meat selections. If at least one of the selections must be vegetarian, how many different dinners could Jane create?
A. 30
B. 40
C. 60
D. 70
E. 80
Mathematics
1 answer:
timama [110]3 years ago
5 0

Answer:

Therefore, Jane can create 80 different dinners.

Step-by-step explanation:

We know that Jane must select three different items for each dinner she will serve. If at least one of the selections must be vegetarian.

The items are to be chosen from among five different vegetarian and four different meat selections.

First we count the number of combinations for one vegetarian dinner and 2 meat dinners.

C_1^5\cdot C_2^4=5\cdot \frac{4!}{2!(4-2)!}=5\cdot 6=30

Now we count the number of combinations for 2 vegetarian dinner and 1 meat dinners.

C_2^5\cdot C_1^4=10\cdot 4=40\\

Now we count the number of combinations for 3 vegetarian dinner.

C_3^5=\frac{5!}{3!(5-3)!}=10\\

We get 30+40+10=80.

Therefore, Jane can create 80 different dinners.

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For this case we must solve each of the equations proposed:

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