How many ways can a committee of 3 people be chosen out of 7 people?

The correct answer is Option (2) → 35

1. The Combination Formula

The number of ways to choose $k$ items from a set of $n$ items is given by:

${n \choose k} = \frac{n!}{k!(n-k)!}$

2. Substitute the Values

Here, $n = 7$ (total people) and $k = 3$ (people to be chosen):

${7 \choose 3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \times 4!}$

3. Simplify the Calculation

Expand the factorials:

${7 \choose 3} = \frac{7 \times 6 \times 5 \times 4!}{ (3 \times 2 \times 1) \times 4!}$

The $4!$ in the numerator and denominator cancel each other out:

${7 \choose 3} = \frac{7 \times 6 \times 5}{6}$

${7 \choose 3} = 7 \times 5 = \mathbf{35}$

Conclusion

There are 35 different ways to form the committee.