6.2 Q-6

Exercise 6.2 — Question 6

Problem

In how many ways can a committee of 3 men and 2 women be formed from a group of 6 men and 5 women?

Key Formula

The number of ways to choose $r$ objects from $n$ distinct objects (where order does not matter) is given by:

$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$

Solution

Step 1: Choose 3 men from 6 men.

$\left(\genfrac{}{}{0ex}{}{6}{3}\right)=\frac{6!}{3!\cdot 3!}=\frac{6\times 5\times 4}{3\times 2\times 1}=20$

Step 2: Choose 2 women from 5 women.

$\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5!}{2!\cdot 3!}=\frac{5\times 4}{2\times 1}=10$

Step 3: Apply the Fundamental Counting Principle.

Since both selections must happen together, multiply the results:

$\text{Total ways}=\left(\genfrac{}{}{0ex}{}{6}{3}\right)\times \left(\genfrac{}{}{0ex}{}{5}{2}\right)=20\times 10=200$

Answer

The committee can be formed in 200 ways.

Concept Reminder

• Use combinations (not permutations) when the order of selection does not matter.
• When two independent selections must both occur, multiply the individual combination counts (Fundamental Counting Principle).
• Real-world applications include forming teams, selecting lottery numbers, and choosing subsets from a population.