Exercise 6.2 — Question 16

Problem Statement

A committee of 5 members is to be formed from 6 men and 4 women. In how many ways can this be done if:

(i) No restriction is imposed (ii) At least 2 women must be included (iii) At most 2 women are included

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Key Formula

The number of ways to choose $r$ items from $n$ distinct items (order does not matter) is:

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

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Solution

Part (i): No Restriction

Choose any 5 members from all $6+4=10$ people:

$\left(\genfrac{}{}{0ex}{}{10}{5}\right)=\frac{10!}{5!\cdot 5!}=\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}=252$

$\boxed{252\text{ ways}}$

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Part (ii): At Least 2 Women

"At least 2 women" means 2, 3, or 4 women on the committee.

$\text{Total}=120+60+6=\boxed{186\text{ ways}}$

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Part (iii): At Most 2 Women

"At most 2 women" means 0, 1, or 2 women on the committee.

$\text{Total}=6+60+120=\boxed{186\text{ ways}}$

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Key Concepts Used

• Combination formula: $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$ — used when order does NOT matter.
• "At least" problems: Sum over all valid cases from the minimum count upward.
• "At most" problems: Sum over all valid cases from zero up to the maximum count.
• Multiplication principle: When selecting from two independent groups (men and women), multiply the number of ways for each group.