Question

There are 8 people on a basketball team, and the coach needs to choose 5 to put into a
game. How many different possible ways can the coach choose a team of 5 if each
person has an equal chance of being selected?

Answer 1

Answer:

There are 56 possible ways for the coach to choose the team.

Step-by-step explanation:

The order in which the players are selected is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

5 players from a set of 8. So

$C_{8,5} = \frac{8!}{5!(8-5)!} = 56$

There are 56 possible ways for the coach to choose the team.

Answer 2

Answer:

56 ways

Step-by-step explanation:

From the question, the formula we can use is the combination formula

The combination formula is given as:

nCr = n!/r! (n-r)!

From the question, n = 8 and r = 5

8C5 = 8!/5! (8-5)!

= (8×7×6×5×4×3×2×1)/ (5×4×3×2×1)(3×2×1)

= 56 ways

Therefore, the different possible ways the coach can choose a team of 5 if each person has an equal chance of being selected is 56 ways.