There are 15 people to choose from for a nine-person batting order in a baseball team. In how many ways can the people be arranged in the lineup?
Is it 1,816,214,400?
2. A basketball team consists of 12 people. How many lineups of 5 people can be selected?
Is it 792?
#1 correct
#2 correct, if you consider all the players as identical.
good work.
To calculate the number of ways a group of people can be arranged, you can use the concept of permutations.
1. For the first question, you have 15 people to choose from for a nine-person batting order. In this case, since the order in which the people are arranged matters (first, second, third, etc.), you can use the formula for permutations:
nPr = n! / (n - r)!
Where n is the total number of options (15 in this case) and r is the number of selections (9 in this case).
Plugging in the values:
15P9 = 15! / (15 - 9)!
= (15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7!) / (6!)
= 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8
= 1,816,214,400
So, your answer is correct. There are 1,816,214,400 ways to arrange the people in a nine-person batting order.
2. For the second question, you have 12 people to choose from for a five-person lineup. Again, since the order matters (first, second, third, etc.), you can use the permutation formula:
12P5 = 12! / (12 - 5)!
= (12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (7 * 6 * 5 * 4 * 3 * 2 * 1)
= 12 * 11 * 10 * 9 * 8
= 79,200
So, your answer is incorrect. There are actually 79,200 ways to select a lineup of 5 people from a group of 12.