6.2 Q-2 | Permutation, Combination And Probability | Mathematics 11

Exercise 6.2 — Question 2

The number of ways to choose $r$ items from $n$ distinct items (without regard to order) is given by:

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

where $n \geq r \geq 0$ .

Problem: In how many ways can a committee of 3 be selected from 8 people?

Solution:

We need to choose 3 people from 8, where order does not matter (a committee is a combination, not a permutation).

$(3 8) = \frac{8 !}{3 ! ( 8 - 3 )!} = \frac{8 !}{3 ! \cdot 5 !}$

$= \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$

Answer: There are 56 ways to form the committee.

Concept	Formula	When to Use
Combination	$(r n) = \frac{n !}{r ! ( n - r )!}$	Order does not matter
Permutation	$P (n, r) = \frac{n !}{( n - r )!}$	Order does matter

Remember: Use combinations when selecting a group, team, committee, or subset where the arrangement does not matter.

Lottery: Choosing 6 numbers from 49 — $(6 49)$ gives the total number of possible tickets.
DNA sequences: Selecting bases to form a codon from available nucleotides.
Cryptography: Choosing a subset of keys or characters for a cipher.