a baseball team has 15 members, 3 of whom always play and bat at the top of the batting order. For the game Sunday, the starting lineup of nine players will be filed out from the remaining members and a batting order will be set. How many possible betting orders can be made?
To find the number of possible batting orders, we need to consider the players who will be playing and batting at the top of the order and the remaining players who will fill the remaining spots in the lineup.
Since 3 players always play and bat at the top of the order, we have 3 fixed positions in the lineup. The remaining 6 positions can be filled by the remaining 12 players.
The number of ways to arrange the 12 remaining players in the lineup can be calculated using the concept of permutations. In this case, we have 12 players and 6 positions to fill. So, we can use the formula for permutations:
nPr = n! / (n - r)!
where n is the total number of objects and r is the number of objects to be selected.
In this case, n = 12 and r = 6.
Substituting the values into the formula:
12P6 = 12! / (12 - 6)!
= 12! / 6!
= (12 * 11 * 10 * 9 * 8 * 7 * 6!) / 6!
= 12 * 11 * 10 * 9 * 8 * 7
Calculating this expression:
12 * 11 * 10 * 9 * 8 * 7 = 66,528
Therefore, there are 66,528 possible batting orders that can be made from the remaining players.