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Bunuel
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Number of ways in which entrees can be chosen = 12C3 = 12!/3!9! =220

Number of ways in which desserts can be chosen = 3C2 = 3

Number of ways in which both entrees AND desserts can be selected = 220*3 = 660

(D) is the correct answer imo

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Bunuel
Kate and Chad are planning their wedding dinner and must select 3 of 12 entrees and 2 of 3 desserts for their guests to be able to choose from. How many different combinations of
offerings are possible?

A. 550
B. 600
C. 620
D. 660
E. 1320

Choice of 3 out of 12 entrees - 12C2

Choice of 2 out of 3 deserts - 3C2

Total Chances = 12C2*3C2 = 660
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IMO D

Ways to select 3 of 12 entrees = 12C3 = 12!/ (3! * 9!) = 220
Ways to select 2 of 3 desserts = 3C2 = 3!/(2! * 1!) = 3

Total combinations of offerings = 220 * 3 = 660
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