If you have 8 outfielders and three starting positions, you have 56 possible starting combinations, right?

Question

I'm writing an article for a school paper. Math has never been my strong suit, and I just want to make sure I don't look too silly when I turn it in. Thanks in advance for the help!

Answer 1

If you definitely want the number of COMBINATIONS, then Yes, 56:

Number of combinations of n things taken r at a time is nCr = n! / [(n-r)! X r!], where "!" is called factorial, such as 3!=3x2x1=6.


n =8
r =3
8C3 = 8! / [5! X 3!] = 56

But combinations suggests it doesn't matter which player plays which outfield position. I think in this case, it does matter which player plays which position. If you agree with me, then you want permutations:

Number of permutations of n things taken r at a time is nPr = n! / (n-r)!

nPr = 8! / 5! = 336

Answer 2

From the way you phrased it, you have 336 possible starting combinations

8 x 7 x 6

You have 8 different possibilities for left field, which leaves 7 possibilities for center field, then 6 for right field

From the way you phrased it, if you had outfielders called A, B,C, then A B C (positioned from left to right in the outfield) is a different starting combination then C B A (but using the same players)

If you want the 3 starting outfielders to be unique regardless of position, then divide 336 by 6 as ABC can be arranged in the outfield in 6 different ways
(3 x 2 x1)

ABC ACB BAC BCA CAB CBA

Which is your answer of 56

Answer 3

Using my calculator, 8 choose 3 is 56, so it looks as if you have the right answer.

In more detail, n choose k = n! / (k! (n-k)!)

Which would be 8! / (3! * 5!)

If you expand the factorials, you will see that everything cancels out except 8*7 which is 56, as you suggested.

Answer 4

I think you need to explain a little more what you are talking about.