In how many ways can a committee of $3$ members be selected from a group of $6$ people?

Solution

1. Since the order of selection does not matter, we use the combination formula $C(n,r)=\frac{n!}{r!(n-r)!}$, where $n=6$ and $r=3$.
2. Substitute the values into the formula: $$\begin{array}{rl}C(6,3)& =\frac{6!}{3!(6-3)!}\\ & =\frac{6\times 5\times 4}{3\times 2\times 1}\end{array}$$.
3. Simplify the expression to find the total number of ways: $\frac{120}{6}=20$.
4. Hence the answer is (B).