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? ​

Answer 1

ermmm...yeah

Explanation:

Since you bring 12 items for 4 dinners, you have a total of 12 items to choose from.

For the first dinner, you need to choose 3 items out of the 12. You can do this in:

12 choose 3 = (12!)/(3!*(12-3)!) = 220 ways

For the second dinner, you have used up 3 items in the first dinner, so you have 9 items left to choose from. You need to choose 3 items out of the 9. You can do this in:

9 choose 3 = (9!)/(3!*(9-3)!) = 84 ways

For the third dinner, you have already used up 6 items, so you have 6 items left to choose from. You need to choose 3 items out of the 6. You can do this in:

6 choose 3 = (6!)/(3!*(6-3)!) = 20 ways

For the fourth dinner, you have already used up 9 items, so you have only 3 items left to choose from. You need to choose all 3 items. You can do this in:

3 choose 3 = (3!)/(3!*(3-3)!) = 1 way

Therefore, you can choose items for the first dinner in 220 ways, for the second dinner in 84 ways, for the third dinner in 20 ways, and for the fourth dinner in 1 way.