Question

There are 10 professors and 20 students. Out of these a committee of 2 professors and 3 students is to be formed. In how many ways can it be done ?

Answer 1

Answer :-

Committee of 2 professors to be formed out of 10 is given by ¹⁰C₂.

¹⁰C₂ = 10! / (2! x 8!) = 45

Committee of 3 students to be formed out of 20 is given by ²⁰C₃.

²⁰C₃= 20!/(17! x 3!) = 1140

So total number of ways in which committee of 2 professors and 3 students is to be formed=  ¹⁰C₂ x ²⁰C₃ = 45 x 1140 = 51300

Answer 2

Given:

The Group of 2 professors to be made of 10 shall be,

¹⁰C₂

The Group of 3 professors to be made of 20 shall be,

²⁰C₃

To Find:

Total number of ways = ?

Solution:

On applying formula in the given expressions, we get

⇒ ¹⁰C₂

⇒  $\frac{10!}{(2!\times 8!)}$

⇒ 45

⇒ ²⁰C₃

⇒ $\frac{20!}{(17!\times 3!)}$

⇒ 1140

So, Total number of ways,

= ¹⁰C₂ × ²⁰C₃

= 45 × 1140

= 51300

Thus, total no. of ways are "51300"