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If there are eleven people on the team and nine positions, how many total arrangements of the players can be made? if amy does not want to play pitcher, then there are now people available to pitch. assuming the pitcher has already been chosen, there are ten remaining players and remaining positions. how many ways are there to arrange the remaining players and positions?

Sagot :

There are 19958400 ways if 9 people are selected for positions and if Amy does not want to play then the number of ways will be 1814400 if done through permutations.

Given there are 11 people and nine positions.

We have to find the number of ways in which arrangement can be done if there are 11 people and 9 positions and the number of ways to arrange if Amy doe not want to play assuming the pitcher has already been chosen.

Permutations helps us to know the number of ways an object can be arranged. It is denoted by n[tex]P_{r}[/tex]=n!/(n-r)!

where n is the number of choices available,

r is the choices.

Given that the team has 11 players while there are only 9 positions available, the number of ways is 11[tex]P_{9}[/tex]=11!/(11-9)!

=11!/2!

=19958400

Now Amy does not want to play, therefore the number of ways the remaining people can be arranged in the place of pitcher is 10. Also the positions has also decreased from 9 to 8.

Thus the number of ways in which remaining people can be arranged are: 10[tex]P_{8}[/tex]=10!/(10-8)!

=10!/2!

=1814400

Hence the number of ways in which 11 people are arranged is 19958400 and the number of ways in which the remaining players are arranged is 1814400.

Learn more about permutation at https://brainly.com/question/1216161

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