At a music festival, there are ten bands scheduled to play, numbered 1 through 10.
a. How many different ways can these bands be arranged to perform?
b. If band 4 is performing first and band 7 last, then how many ways can their appearances be scheduled?
a. There are $\square$ different ways to arrange the bands.
(Simplify your answer.)
b. If band 4 is performing first and band 7 last, there are $\square$ different ways to arrange the bands.
(Simplify your answer.)
Answer
The number of permutations of 8 bands is \(8!\), which simplifies to \(\boxed{40320}\) different ways.
Step 1 ：There are 10 bands scheduled to perform at a music festival, numbered 1 through 10.
Step 2 ：To find the number of different ways these bands can be arranged to perform, we need to calculate the number of permutations of 10 bands.
Step 3 ：The number of permutations of 10 bands is \(10!\), which simplifies to \(\boxed{3628800}\) different ways.
Step 4 ：If band 4 is performing first and band 7 is performing last, we need to find the number of permutations of the remaining 8 bands.
Step 5 ：The number of permutations of 8 bands is \(8!\), which simplifies to \(\boxed{40320}\) different ways.
