Question

a club with eleven members is to choose three officers: president, vice - president, and secretary - treasurer. if each office is to be held by one person and no person can hold more than one office, in how many ways can those offices be filled?
ways

Explanation:

Step1: Identify the problem type

This is a permutation problem since the order of selection (president, vice - president, secretary - treasurer) matters. The formula for permutations of \(n\) objects taken \(r\) at a time is \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 11\) (total number of club members) and \(r=3\) (number of offices to be filled).

Step2: Apply the permutation formula

We know that \(n = 11\) and \(r=3\). Substitute these values into the formula \(P(n,r)=\frac{n!}{(n - r)!}\).
First, calculate \(n-r=11 - 3=8\).
Then, \(P(11,3)=\frac{11!}{8!}\). Since \(n!=n\times(n - 1)\times\cdots\times1\), we can write \(11! = 11\times10\times9\times8!\).
So, \(\frac{11!}{8!}=\frac{11\times10\times9\times8!}{8!}\). The \(8!\) terms in the numerator and denominator cancel out.
Now, calculate \(11\times10\times9 = 990\).

Answer:

990