Exercise 6.2 — Question 20

Problem Statement

How many committees of 5 members can be formed from a group of 8 men and 6 women if the committee must contain at least 2 women?

Note: If the original problem statement differs, replace the above with the exact textbook question. The solution method below applies to the standard FBISE Exercise 6.2 Q20 format.

Key Formula

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

$(r n) = \frac{n !}{r ! ( n - r )!}$

Solution

We need committees of 5 members from 8 men and 6 women with at least 2 women.

"At least 2 women" means: exactly 2, 3, 4, or 5 women.

Case 1: Exactly 2 women, 3 men

$(2 6) \times (3 8) = 15 \times 56 = 840$

Case 2: Exactly 3 women, 2 men

$(3 6) \times (2 8) = 20 \times 28 = 560$

Case 3: Exactly 4 women, 1 man

$(4 6) \times (1 8) = 15 \times 8 = 120$

Case 4: Exactly 5 women, 0 men

$(5 6) \times (0 8) = 6 \times 1 = 6$

Total

$840 + 560 + 120 + 6 = 1526$

Key Concepts Used

Combination formula $(r n)$ : used when order does not matter.
Addition Principle: when an event can occur in mutually exclusive ways, add the counts.
Multiplication Principle: when two independent selections are made, multiply the counts.
"At least" condition: handled by summing all valid cases (or using complement method).

Complement Method (Alternative)

Total committees of 5 from 14 people: $(5 14) = 2002$

Committees with 0 women (all men): $(5 8) = 56$

Committees with exactly 1 woman: $(1 6) \times (4 8) = 6 \times 70 = 420$

At least 2 women $= 2002 - 56 - 420 = 1526$ ✓

Real-World Connection (SLO M-11-C-04)

Combinations appear in:

Lottery odds: choosing 6 numbers from 49 gives $(6 49) = 13, 983, 816$ possible tickets.
DNA sequences: selecting bases for a codon from available nucleotides.
Team selection: choosing project teams, sports squads, or committee members.

Exercise 6.2 — Question 20

Problem Statement

How many committees of 5 members can be formed from a group of 8 men and 6 women if the committee must contain at least 2 women?

Note: If the original problem statement differs, replace the above with the exact textbook question. The solution method below applies to the standard FBISE Exercise 6.2 Q20 format.

Key Formula

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

$(r n) = \frac{n !}{r ! ( n - r )!}$

Solution

We need committees of 5 members from 8 men and 6 women with at least 2 women.

"At least 2 women" means: exactly 2, 3, 4, or 5 women.

Case 1: Exactly 2 women, 3 men

$(2 6) \times (3 8) = 15 \times 56 = 840$

Case 2: Exactly 3 women, 2 men

$(3 6) \times (2 8) = 20 \times 28 = 560$

Case 3: Exactly 4 women, 1 man

$(4 6) \times (1 8) = 15 \times 8 = 120$

Case 4: Exactly 5 women, 0 men

$(5 6) \times (0 8) = 6 \times 1 = 6$

Total

$840 + 560 + 120 + 6 = 1526$

Key Concepts Used

Combination formula $(r n)$ : used when order does not matter.
Addition Principle: when an event can occur in mutually exclusive ways, add the counts.
Multiplication Principle: when two independent selections are made, multiply the counts.
"At least" condition: handled by summing all valid cases (or using complement method).

Complement Method (Alternative)

Total committees of 5 from 14 people: $(5 14) = 2002$

Committees with 0 women (all men): $(5 8) = 56$

Committees with exactly 1 woman: $(1 6) \times (4 8) = 6 \times 70 = 420$

At least 2 women $= 2002 - 56 - 420 = 1526$ ✓

Real-World Connection (SLO M-11-C-04)

Combinations appear in:

Lottery odds: choosing 6 numbers from 49 gives $(6 49) = 13, 983, 816$ possible tickets.
DNA sequences: selecting bases for a codon from available nucleotides.
Team selection: choosing project teams, sports squads, or committee members.

6.2 Q-20

Exercise 6.2 — Question 20

Problem Statement

Key Formula

Solution

Case 1: Exactly 2 women, 3 men

Case 2: Exactly 3 women, 2 men

Case 3: Exactly 4 women, 1 man

Case 4: Exactly 5 women, 0 men

Total

Key Concepts Used

Complement Method (Alternative)

Real-World Connection (SLO M-11-C-04)

6.2 Q-20

Exercise 6.2 — Question 20

Problem Statement

Key Formula

Solution

Case 1: Exactly 2 women, 3 men

Case 2: Exactly 3 women, 2 men

Case 3: Exactly 4 women, 1 man

Case 4: Exactly 5 women, 0 men

Total

Key Concepts Used

Complement Method (Alternative)

Real-World Connection (SLO M-11-C-04)