Question

A baseball team has 15 players, but there are only 9 positions. How many different choices are there for how to put 9 players on the field?

Answer 1

The number of different choices there are to put 9 players out of 15 on the field is 1816214400

Since we require the different ways of choosing 15 players out of 9, this is a permutation question since order matters. Note- each player is unique and cannot be chosen twice.

Number of Permutations

So, we have 15 choices for the first player, 14 choices for the second player, 13 choices for the third player, 12 choices for the fourth players, 11 choices for the fifth player, 10 choices for the sixth player, 9 choices for the seventh player, 8 choices for the eight player and 7 choices for the ninth player.

Number of ways of choosing 9 players out of 15

So, our number of permutations is 9 players to be selected out of 15 which is ¹⁵P₉ = 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7

= 1816214400 choices

So, the number of different choices there are to put 9 players out of 15 on the field is 1816214400

Learn more about permutations here:

https://brainly.com/question/6774850