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?
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 $220 \times 84 \times 20 \times 1 = 739,200$.
To find the number of ways to choose 3 items for each dinner, we can use combinations.
For the first dinner, you can choose 3 items out of the 12 available. The number of ways to choose 3 items from 12 is calculated as:
C(12, 3) = 12! / (3!(12-3)!) = 220
So, there are 220 ways to choose 3 items for the first dinner.
Similarly, for the second dinner, there will now be 9 items left to choose from (since you have already used 3 items for the first dinner). So, the number of ways to choose 3 items for the second dinner is:
C(9, 3) = 9! / (3!(9-3)!) = 84
For the third dinner, there will be 6 items left to choose from, and the calculation is:
C(6, 3) = 6! / (3!(6-3)!) = 20
Finally, for the fourth dinner, there will be 3 items left to choose from, and the calculation is:
C(3, 3) = 3! / (3!(3-3)!) = 1
Therefore, there are 220 ways to choose the 3 items for the first dinner, 84 ways for the second dinner, 20 ways for the third dinner, and 1 way for the fourth dinner.