There are eleven people on a softball team and nine different positions. work through the questions to determine how many ways a coach can choose the players for the positions if amy does not want to pitch

19958400

10

8

1814400

Step-by-step explanation:

* Lets explain how to solve the problem

- Permutation is the act of arranging the members into order

- The number of permutations of n members taken r at a time is

  denoted by nPr

- nPr = n!/(n - r)! , where n! means n(n - 1)(n - 2)(n - 3) × 1

* Lets solve the problem

- There are eleven people on a softball team

∴ n = 11

- There are nine different positions

∴ r = 9

∴ The number of ways can be made = 11P9

∵ 11P9 = 11!/(11 - 9)! = 11!/2! = 19958400

* The number of total arrangements of the player is 19958400

- If Amy does not want to play pitcher

∵ They are 11 players

∴ The number of players can be pitch = 11 - 1 = 10

* If Amy does not want to play pitcher, then there are now 10 people

 available to pitch

- Assuming the pitcher has already be chosen

∴ There are 10 remaining players

∵ The positions are 9

∴ The remaining positions = 9 - 1 = 8

* There are 8 remaining positions

- Lets find how many ways to arrange the remaining positions

∵ n = 10

∵ r = 8

∴ 10P8 = 10!/(10 - 8)! = 10!/2! = 1814400

* The number of ways to arrange the remaining players is 1814400

im pretty sure it would be 5 am, hope that , and give brainliest and guys if this is right then press the button and give 5 stars

yes   because as the functions are graphed they eventually will intersect unless otherwise

step-by-step explanation: