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

The problem requires choosing 3 people out of 7 people for a committee.

In a committee, order doesn't matter. Choosing Ram, Sita, and Geeta is the same as choosing Geeta, Ram, and Sita. This means combinations are used.

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When choosing $r$ things from $n$ things where order doesn't matter:

$C(n,r) = \dfrac{n!}{r! \times (n-r)!}$

where $n!$ (n factorial) $= n \times (n-1) \times (n-2) \times ... \times 1$

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For this problem:

• $n = 7$ (total people)
• $r = 3$ (people to choose)

$C(7,3) = \dfrac{7!}{3! \times 4!}$

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Instead of calculating $7! = 5040$, the expression can be simplified:

$C(7,3) = \dfrac{7 \times 6 \times 5 \times 4!}{3! \times 4!}$

The $4!$ cancels from numerator and denominator:

$= \dfrac{7 \times 6 \times 5}{3!}$

$= \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}$

$= \dfrac{210}{6}$

$= 35$

Therefore, the committee can be chosen in 35 ways.