How many combinations of 5 players can be created from a group of 9 players?

How do I solve this problem?

Thanks for any help!

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There is a formula for this involving *factorials. *A factorial is the starting number (A) times (A-1) times (A-2) etc down to A times 1. The symbol for factorial is "!" We use factorials for these problems, because they represent the number of combinations we can get from a group of numbers.

Let's say the total number in the group is X. The size of each combination is Y.

The formula for the number of combinations is:

X! / Y! (x-y)!

In your problem, X=9 and y=5. So, the formula works out to:

9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 / (5 x 4 x 3 x 2 x 1) times (4 x 3 x 2 x 1)

9! = 362880

5! = 120

4! = 24

Substituting these amounts, we get:

362880 / 120 x 24

or

362880 / 2880

or

126

You know there is a group of nine players so lets name it n players in total. You also know the size of each possible combination is 5 players, so lets give in r players in each combination.

For this problem, we are using a formula called a binomial coefficient and is denoted by nCr, which is n choose r.

The general equation is n!/ r!(n-r)!

n!=9!

= n(n-1)(n-2)(n-3)....

= 9*8*7*6*5*4*3*2*1

= 362880

r!=5!

= r(r-1)(r-2)(r-3)...

= 5*4*3*2*1

= 120

(n-r)!= (9-5)!

= 4!

= 4*3*2*1

= 24

so, combination is= 362880/ 120*24

= 362880/ 2880

= 126 possibilities

uhh simple it's just 9C5= 9!/5!4! != factorial i.e 5! = 5 x 4 x 3 x 2 x 1, 3!= 3 x 2 x 1

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