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 the 3 items for the first dinner? For the second? For the third? For the fourth?

Answer 1

For the first dinner, you have to choose 3 items out of 12. This can be done in $\binom{12}{3} = 220$ ways.

After you've chosen the 3 items for the first dinner, there are only 9 items left. For the second dinner, you have to choose 3 items out of these 9. This can be done in $\binom{9}{3} = 84$ ways.

Similarly, for the third dinner, you have to choose 3 items out of the remaining 6. This can be done in $\binom{6}{3} = 20$ ways.

Finally, for the fourth dinner, you have to choose 3 items out of the last 3. This can be done in only $\binom{3}{3} = 1$ way.

Therefore, the total number of ways to choose the items for all 4 dinners is $220 \times 84 \times 20 \times 1 = 739,200$.