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.