6.2 Q-18 | Permutation, Combination And Probability | Mathematics 11

Exercise 6.2 — Question 18

Problem Statement

How many committees of 5 members can be formed from 8 men and 6 women if the committee must consist of 3 men and 2 women?

Note: If the original problem statement differs, replace accordingly. The solution method below applies to any combination-selection problem of this type.

Key Formula

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

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

Solution

Step 1: Choose 3 men from 8 men.

$(3 8) = \frac{8 !}{3 ! \cdot 5 !} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Step 2: Choose 2 women from 6 women.

$(2 6) = \frac{6 !}{2 ! \cdot 4 !} = \frac{6 \times 5}{2 \times 1} = 15$

Step 3: Apply the Fundamental Counting Principle.

Since the two selections are independent, multiply the results:

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

Answer: $840$ committees can be formed.

Concept Summary

Concept	Description
Combination $(r n)$	Selection of $r$ items from $n$ where order does not matter
Fundamental Counting Principle	If event A can occur in $m$ ways and event B in $n$ ways, both together occur in $m \times n$ ways
When to use combinations	Forming committees, teams, groups — any selection without arrangement

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

Combinations appear in many real-world contexts:

Lottery odds: Choosing 6 numbers from 49 gives $(6 49) = 13, 983, 816$ possible tickets.
DNA sequences: Selecting bases to form codons uses combinatorial counting.
Team selection: Choosing project teams, sports squads, or committee members.

Exercise 6.2 — Question 18

Problem Statement

How many committees of 5 members can be formed from 8 men and 6 women if the committee must consist of 3 men and 2 women?

Note: If the original problem statement differs, replace accordingly. The solution method below applies to any combination-selection problem of this type.

Key Formula

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

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

Solution

Step 1: Choose 3 men from 8 men.

$(3 8) = \frac{8 !}{3 ! \cdot 5 !} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Step 2: Choose 2 women from 6 women.

$(2 6) = \frac{6 !}{2 ! \cdot 4 !} = \frac{6 \times 5}{2 \times 1} = 15$

Step 3: Apply the Fundamental Counting Principle.

Since the two selections are independent, multiply the results:

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

Answer: $840$ committees can be formed.

Concept Summary

Concept	Description
Combination $(r n)$	Selection of $r$ items from $n$ where order does not matter
Fundamental Counting Principle	If event A can occur in $m$ ways and event B in $n$ ways, both together occur in $m \times n$ ways
When to use combinations	Forming committees, teams, groups — any selection without arrangement

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

Combinations appear in many real-world contexts:

Lottery odds: Choosing 6 numbers from 49 gives $(6 49) = 13, 983, 816$ possible tickets.
DNA sequences: Selecting bases to form codons uses combinatorial counting.
Team selection: Choosing project teams, sports squads, or committee members.