Answer and Step-by-step explanation:
This is a problem of permutations. Let’s solve each part separately:
(a) If there are no restrictions, then any of the 9 players can bat first, any of the remaining 8 players can bat second, any of the remaining 7 players can bat third, and so on. So the total number of possible batting orders is 9! = 362880.
(b) If the pitcher must bat last, then there are 8 choices for the first batter, 7 choices for the second batter, and so on, until there is only one choice left for the eighth batter. The pitcher must bat ninth. So the total number of possible batting orders is 8! = 40320.
© If the pitcher must bat last, the catcher eighth, and the shortstop first, then there are 6 choices for the second batter, 5 choices for the third batter, and so on, until there is only one choice left for the seventh batter. So the total number of possible batting orders is 6! = 720.