Question

a coach chooses six out of eight players to go to a skills workshop. if order does not matter, in how many ways can he choose the players for the workshop? 6 8 56 28

Explanation:

Step1: Identify the problem type

This is a combination problem since the order of choosing the players does not matter. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n \) is the total number of items, and \( k \) is the number of items to choose. Here, \( n = 8 \) (total players) and \( k=6 \) (players to choose).

Step2: Apply the combination formula

First, substitute \( n = 8 \) and \( k = 6 \) into the formula:

$$ C(8,6)=\frac{8!}{6!(8 - 6)!}=\frac{8!}{6!2!} $$

We know that \( n!=n\times(n - 1)\times\cdots\times1 \), so \( 8! = 8\times7\times6! \) and \( 2! = 2\times1 \). Substitute these into the formula:

$$ C(8,6)=\frac{8\times7\times6!}{6!\times2\times1} $$

The \( 6! \) terms cancel out, leaving:

$$ C(8,6)=\frac{8\times7}{2\times1}=\frac{56}{2}=28 $$

Answer:

28 (the option with the number 28)