if 10 students enter projects in the art show, how many different ways can they come in 1st, 2nd, and 3rd place?
would you go 10^3
OR 10TIMES 3?
This is not a trig question.
Neither of those answers is correct.
After making one of ten possible choices for first place, there are 9 possible choices for second place, etc.
The answer is 10*9*8 = 720
To find the number of different ways the 10 students can come in 1st, 2nd, and 3rd place in the art show, you need to use the concept of permutations.
The number of ways the 10 students can come in 1st place is 10 (since any of the 10 students can come in 1st).
For each of these 10 choices for 1st place, there are 9 remaining students who can come in 2nd place (since one student already took 1st place). So, the number of ways the 10 students can come in both 1st and 2nd place is 10 * 9 = 90.
Finally, for each of these 90 choices for 1st and 2nd place, there are 8 remaining students who can come in 3rd place. So, the total number of ways the 10 students can come in 1st, 2nd, and 3rd place is 10 * 9 * 8 = 720.
Therefore, the correct answer is 720, which is obtained by multiplying the number of choices for each position (10 for 1st place, 9 for 2nd place, and 8 for 3rd place) together.