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

Question:

There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees:

(i) a particular professor is included.

(ii) a particular student is included.

(iii) a particular student is excluded.

Solution:

Clearly, 2 professors and 3 students are selected out of 10 professors and 20 students, respectively.

Required number of ways $={ }^{10} C_{2} \times{ }^{20} C_{3}=\frac{10}{2} \times \frac{9}{1} \times \frac{20}{3} \times \frac{19}{2} \times \frac{18}{1}=51300$

(i) If a particular professor is included, it means that 1 professor is selected out of the remaining 9 professors.

Required number of ways $={ }^{20} C_{3} \times{ }^{9} C_{1}=\frac{20}{3} \times \frac{19}{2} \times \frac{18}{1} \times \frac{9}{1}=10260$

(ii) If a particular student is included, it means that 2 students are selected out of the remaining 19 students.

Required number of ways $={ }^{19} C_{2} \times{ }^{10} C_{2}=\frac{19}{2} \times \frac{18}{1} \times \frac{10}{2} \times \frac{9}{1}=7695$

(iii) If a particular student is excluded, it means that 3 students are selected out of the remaining 19 students.

Required number of ways $={ }^{19} C_{3} \times{ }^{10} C_{2}=\frac{19}{3} \times \frac{18}{2} \times \frac{17}{1} \times \frac{10}{2} \times \frac{9}{1}=43605$