Question

a company is sending 5 of its 10 members to a conference, how many different sets of 5 members can be seated?

Answer 1

The answer to this is \[ \dbinom {10}{5}, \]where \[ \dbinom {n}{k} \]is the number of ways to select k objects out of n and is defined as \[ \frac {n!}{k! \cdot (n-k)!}, \]so the answer is \[ \frac {10!}{5! 5!} = \frac {10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \boxed{252}. \]