How many ways can a baseball coach arrange the order of nine batters if there are 15 players on the team?
Well, they’re going to be nine batters, so batters in position one through nine. For the first batter, the coach will have the choice from all 15 players. For the second batter, he’ll only be able to choose from 14 of the players because one of them will already be in the first position. For the third position, there will be 13 available players then 12, 11, 10, nine, eight, and seven to choose from.
To find the total number of ways that the baseball coach can arrange the order, we need to multiply each choice. 15 times 14 times 13 all the way down to seven. The baseball coach would have 1816214400 different ways to arrange his 15 players taking nine at a time.
This we can represent with a permutation. A permutation of 𝑛, 𝑟 this is the formula for permutations without repetition. The player who bats first could not also bat second. In this equation, we substitute the number of players for the letter 𝑛 and the number of batters for the letter 𝑟: 15 factorial over 15 minus nine factorial. 15 factorial looks like this. 15 minus nine equals six. So our denominator is six factorial.
We’ve expanded 15 factorial and six factorial. What we see is that six factorial in the denominator cancels out six times five times four times three times two times one in the numerator, which is the same expression we calculated earlier.
15 factorial over six factorial equals 1816214400.