The coach of a soccer team is holding tryouts and can take only
3
more players for the team. There are
5
players trying out. How many different groups of
3
players could possibly be chosen?
5 choose 3 equals 5!/3!2!=5x4/2=10
☺☺☺☺
1, 2, 3, 4, 5.
(1). 1, 2, 3.
(2). 2, 3, 4.
(3). 3, 4, 5.
(4). 4, 5, 1.
(5). 5, 1, 2.
5 Groups.
(6). 1. 3, 5.
(7). 2, 4, 1.
(8). 3, 5, 2.
(9). 4, 1, 3.
(10). 5, 2, 4.
Total = 10 Groups.
The coach of a soccer team is holding tryouts and can take only 2 more players for the team. There are 6 players trying out. How many different groups of players could possibly be chosen?
To find the number of different groups of 3 players that could be chosen from a pool of 5 players, we can use combinations.
A combination is a selection of items without regard to the order. The formula for combinations is:
C(n, r) = n! / (r!(n-r)!)
where C(n, r) denotes the number of combinations of choosing r items from a pool of n items, and the exclamation mark denotes factorial.
In this case, there are 5 players and the coach needs to choose 3 players. Using the combination formula, we can calculate C(5, 3) as follows:
C(5, 3) = 5! / (3!(5-3)!)
= 5! / (3!2!)
= (5 * 4 * 3!) / (3! * 2 * 1)
= 5 * 4 / (2 * 1)
= 10
Therefore, there are 10 different groups of 3 players that could possibly be chosen from the 5 players trying out.