There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed.

Given:

Total number of professor = 10

Total number of students = 20

Number of ways = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

= (¹⁰C₂) × (²⁰C₃)

By using the formula,

ⁿC_r = \( \dfrac{ n!}{r!(n – r)!}\)

¹⁰C₂ × ²⁰C₃ = \(\dfrac{ 10!}{2!(10 – 2)!} ×\dfrac{ 20!}{3!(20-3)!}\)

= 5×9 × 10×19×6

= 45 × 1140

= 51300 ways

(i) a particular professor is included.

Number of ways = (choosing 1 professor out of 9 professors) × (choosing 3 students out of 20 students)

= ⁹C₁ × ²⁰C₃

By using the formula,

ⁿC_r =\( \dfrac{ n!}{r!(n – r)!}\)

⁹C₁ × ²⁰C₃ =\( \dfrac{9!}{1!(9 – 1)! }× \dfrac{20!}{3!(20-3)!}\)

= 9 × 10×19×6

= 10260 ways

(ii) a particular student is included.

Number of ways = (choosing 2 professors out of 10 professors) × (choosing 2 students out of 19 students)

= ¹⁰C₂ × ¹⁹C₂

By using the formula,

ⁿC_r = \( \dfrac{ n!}{r!(n – r)!}\)

¹⁰C₂ × ¹⁹C₂ = \(\dfrac{ 10!}{2!(10 – 2)!} × \dfrac{19!}{2!(19-2)! }\)

= 5×9 × 19×9

= 45 × 171

= 7695 ways

(iii) a particular student is excluded.

Number of ways = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 19 students)

= ¹⁰C₂ × ¹⁹C₃

By using the formula,

ⁿC_r =\( \dfrac{ n!}{r!(n – r)!}\)

¹⁰C₂ × ¹⁹C₃ = \(\dfrac{ 10!}{2!(10 – 2)!} × \dfrac{19!}{3!(19-3)!}\)

= 5×9 × 19×3×17

= 45 × 969

= 43605 ways

The required no. of ways are 51300, 10260, 7695, 43605.