The coach of a soccer team is holding tryouts and can take only

2
more players for the team. There are
8
players trying out. How many different groups of
2
players could possibly be chosen?

What is C(8,2) or 8!/(6!2!) ??

To find the number of different groups of 2 players that could be chosen from a pool of 8 players, we can use the combination formula.

The combination formula is given by:

C(n, r) = n! / (r! * (n-r)!)

Where n is the total number of items, and r is the number of items chosen.

In this case, we have 8 players (n) and we need to choose 2 players (r), so we can calculate it as:

C(8, 2) = 8! / (2! * (8-2)!)

Simplifying the equation:

C(8, 2) = 8! / (2! * 6!)

Now, let's calculate each factorial:

8! = 8 * 7 * 6! = 40320
2! = 2 * 1 = 2
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

Substituting these values back into the equation:

C(8, 2) = 40320 / (2 * 720)

Now we can simplify by performing the calculations:

C(8, 2) = 40320 / 144

Finally, we get the result:

C(8, 2) = 280

Therefore, there are 280 different groups of 2 players that could possibly be chosen from the pool of 8 players.