Question

On a camping trip you bring 12 items for 4 dinners. For each dinner you use 3 items. In how many ways can you choose items for the first dinner? for the second? for the third? for the fourth?

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Answer 1

• 220 options, 84 options, 20 options, 1 option

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