6.2 Q-12 | Permutation, Combination And Probability | Mathematics 11

Exercise 6.2 — Question 12

In how many ways can a committee of 3 boys and 2 girls be formed from a group of 6 boys and 5 girls?

(Standard FBISE Exercise 6.2 Q12 — selecting a committee using combinations)

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

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

Step 1: Choose 3 boys from 6 boys.

$^{6} C_{3} = \frac{6 !}{3 ! ( 6 - 3 )!} = \frac{6 !}{3 ! \cdot 3 !} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

Step 2: Choose 2 girls from 5 girls.

$^{5} C_{2} = \frac{5 !}{2 ! ( 5 - 2 )!} = \frac{5 !}{2 ! \cdot 3 !} = \frac{5 \times 4}{2 \times 1} = 10$

Step 3: Apply the Fundamental Counting Principle.

Since the two selections are independent, multiply the results:

$Total ways =^{6} C_{3} \times^{5} C_{2} = 20 \times 10 = 200$

A committee is an unordered selection — the order in which members are chosen does not matter. Therefore we use $^{n} C_{r}$ , not $^{n} P_{r}$ .

This type of problem models real-world selection tasks such as: