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Sagot :
Answer:
- 220 options, 84 options, 20 options, 1 option
Step-by-step explanation:
The first dinner
Combinations of 3 out of 12 items
- 12C3 = 12!/(12-3)!3! = 12!/9!3! = 12*11*10/2*3 = 220
The second dinner
Combinations 3 out of remaining 9 items
- 9C3 = 9!/(9 - 3)!3! = 9!/6!3! = 9*8*7/2*3 = 84
The third dinner
Combinations of 3 out of remaining 6 items
- 6C3 = 6!/(6 - 3)!3! = 6!/3!3! = 6*5*4/2*3 = 20
The fourth dinner
Combinations of 3 out of 3 remaining items
- 3C3 = 3!/(3 - 3)!3! = 3!/0!3! = 1
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