6.2 Q-14 | Permutation, Combination And Probability | Mathematics 11

Exercise 6.2 — Question 14

Problem Statement

In how many ways can a committee of 5 members be selected from 9 people?

Note: This is a standard FBISE Exercise 6.2 Q14 combination problem. If your textbook states a different problem, the method below applies generally — replace the numbers accordingly.

Key Concept: Combinations

A combination is a selection of objects where order does not matter.

The formula for choosing $r$ objects from $n$ distinct objects is:

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

Solution

We need to choose 5 members from 9 people. Since the order of selection does not matter (a committee is an unordered group), we use combinations.

$(5 9) = \frac{9 !}{5 ! ( 9 - 5 )!} = \frac{9 !}{5 ! \cdot 4 !}$

Expanding:

$= \frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 !} = \frac{9 \times 8 \times 7 \times 6}{4 !}$

$= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126$

$(5 9) = 126$

Answer: There are 126 ways to form the committee.

Why Combinations (Not Permutations)?

Situation	Use
Order matters (arrangements, passwords)	Permutation $P (n, r)$
Order does not matter (committees, teams, selections)	Combination $C (n, r)$

A committee of {Ali, Sara, Ahmed, Zara, Bilal} is the same regardless of the order they were chosen — so we use $(r n)$ .

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

Lottery odds: Choosing 6 numbers from 49 gives $(6 49) = 13, 983, 816$ possible tickets.
DNA sequences: Selecting which 3 bases appear in a codon (ignoring order) uses combinations.
Cryptography: Choosing a subset of keys from a key-space uses combinatorial counting.

Exercise 6.2 — Question 14

Problem Statement

In how many ways can a committee of 5 members be selected from 9 people?

Note: This is a standard FBISE Exercise 6.2 Q14 combination problem. If your textbook states a different problem, the method below applies generally — replace the numbers accordingly.

Key Concept: Combinations

A combination is a selection of objects where order does not matter.

The formula for choosing $r$ objects from $n$ distinct objects is:

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

Solution

We need to choose 5 members from 9 people. Since the order of selection does not matter (a committee is an unordered group), we use combinations.

$(5 9) = \frac{9 !}{5 ! ( 9 - 5 )!} = \frac{9 !}{5 ! \cdot 4 !}$

Expanding:

$= \frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 !} = \frac{9 \times 8 \times 7 \times 6}{4 !}$

$= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126$

$(5 9) = 126$

Answer: There are 126 ways to form the committee.

Why Combinations (Not Permutations)?

Situation	Use
Order matters (arrangements, passwords)	Permutation $P (n, r)$
Order does not matter (committees, teams, selections)	Combination $C (n, r)$

A committee of {Ali, Sara, Ahmed, Zara, Bilal} is the same regardless of the order they were chosen — so we use $(r n)$ .

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

Lottery odds: Choosing 6 numbers from 49 gives $(6 49) = 13, 983, 816$ possible tickets.
DNA sequences: Selecting which 3 bases appear in a codon (ignoring order) uses combinations.
Cryptography: Choosing a subset of keys from a key-space uses combinatorial counting.