A baseball team's roster consists of 25 players with 13 position players and the rest being pitchers. The team's batting order for a particular game consists of 9 players. The first 8 places in the batting order are taken by position players, The last place is taken by a pitcher. How many batting orders are possible? Use the formula for permutations to find your answer. 15,444 2,042,975 \( 51,891,840 \) \( 622,702,080 \)
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Beyond the Answer
To calculate the number of possible batting orders, we need to arrange the first 8 position players and then choose 1 pitcher for the last spot. First, we can choose 8 position players from 13. The number of ways to choose 8 players is given by the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \): \[ C(13, 8) = \frac{13!}{8!5!} = 1,287. \] Next, we arrange the 8 chosen position players, which can be done in \( 8! = 40,320 \) ways. For the pitcher, we have 12 options (since there are 12 pitchers left). So, the total number of batting orders is: \[ C(13, 8) \times 8! \times 12 = 1,287 \times 40,320 \times 12 = 622,702,080. \] Hence, the answer is \( 622,702,080 \).