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
In baseball, a team has 15 players on its roster, but there are only 9 positions on the field. This means that there are many different ways to arrange the players on the field. The number of different choices is a combinatorial problem, and it can be solved using the combination formula.
The combination formula is:
nCr = n! / (r! * (n-r)!)
where:
- n is the total number of items
- r is the number of items to choose
- n! is the factorial of n
- r! is the factorial of r
- (n-r)! is the factorial of (n-r)
In this case, we have 15 players and we need to choose 9 of them. So, we can plug these values into the combination formula to get the number of different choices:
15C9 = 15! / (9! * (15-9)!)
15C9 = 5005
So, there are 5005 different ways to arrange 9 players on the field.
Related Questions
- How many different ways can you arrange 10 players on the field?
- 252
- How many different ways can you arrange 12 players on the field?
- 924
- How many different ways can you arrange 14 players on the field?
- 3003
- How many different ways can you arrange 16 players on the field?
- 10010
- How many different ways can you arrange 18 players on the field?
- 30030
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