Exercise 6.2 — Question 15

Problem

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

(a) exactly 2 women (b) at least 2 women (c) at most 2 women

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Key Formula

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

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

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Solution

Part (a): Exactly 2 women

We need exactly 2 women from 4, and exactly 3 men from 6.

$\text{Ways}=\left(\genfrac{}{}{0ex}{}{4}{2}\right)\times \left(\genfrac{}{}{0ex}{}{6}{3}\right)$

$=\frac{4!}{2!2!}\times \frac{6!}{3!3!}$

$=6\times 20=\boxed{120}$

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Part (b): At least 2 women

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

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

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Part (c): At most 2 women

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

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

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Key Takeaways

• Exactly $k$: select precisely $k$ from one group and the remainder from the other.
• At least $k$: sum all cases where the count is $\ge k$.
• At most $k$: sum all cases where the count is $\le k$.
• Always verify: (at least 2) + (at most 1) should equal the total number of unrestricted committees.

$\text{Total unrestricted}=\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$

$186+(6+60)=186+66=252✓$