6.2 Q-9 | Permutation, Combination And Probability | Mathematics 11

Exercise 6.2 — Question 9

Problem

How many committees of 5 members can be formed from a group of 8 people?

Note: This is a combination problem because the order of selection does not matter — a committee is the same regardless of the order members are chosen.

Key Formula

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

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

Solution

We need to choose $r = 5$ members from $n = 8$ people.

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

Expanding:

$= \frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 !} = \frac{8 \times 7 \times 6}{3 !} = \frac{336}{6} = 56$

Answer

$(5 8) = 56 committees$

Key Concept

Use combinations (not permutations) when the order of selection does not matter.
$(r n) = (n - r n)$ — note that $(5 8) = (3 8) = 56$ , which provides a useful shortcut.

Exercise 6.2 — Question 9

Problem

How many committees of 5 members can be formed from a group of 8 people?

Note: This is a combination problem because the order of selection does not matter — a committee is the same regardless of the order members are chosen.

Key Formula

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

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

Solution

We need to choose $r = 5$ members from $n = 8$ people.

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

Expanding:

$= \frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 !} = \frac{8 \times 7 \times 6}{3 !} = \frac{336}{6} = 56$

Answer

$(5 8) = 56 committees$

Key Concept

Use combinations (not permutations) when the order of selection does not matter.
$(r n) = (n - r n)$ — note that $(5 8) = (3 8) = 56$ , which provides a useful shortcut.