**Question:**

In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

**Solution:**

It is given that the question paper consists of 12 questions divided into two parts – Part I and Part II, containing 5 and 7 questions, respectively.

A student has to attempt 8 questions, selecting at least 3 from each part.

This can be done as follows.

(a) 3 questions from part I and 5 questions from part II

(b) 4 questions from part I and 4 questions from part II

(c) 5 questions from part I and 3 questions from part II

3 questions from part I and 5 questions from part II can be selected in ${ }^{5} \mathrm{C}_{3} \times{ }^{7} \mathrm{C}_{5}$ ways.

4 questions from part I and 4 questions from part II can be selected in ${ }^{5} \mathrm{C}_{4} \times{ }^{7} \mathrm{C}_{4}$ ways.

5 questions from part I and 3 questions from part II can be selected in ${ }^{5} \mathrm{C}_{5} \times{ }^{7} \mathrm{C}_{3}$ ways.

Thus, required number of ways of selecting questions

$={ }^{5} \mathrm{C}_{3} \times{ }^{7} \mathrm{C}_{5}+{ }^{5} \mathrm{C}_{4} \times{ }^{7} \mathrm{C}_{4}+{ }^{5} \mathrm{C}_{5} \times{ }^{7} \mathrm{C}_{3}$

$=\frac{5 !}{2 ! 3 !} \times \frac{7 !}{2 ! 5 !}+\frac{5 !}{4 ! 1 !} \times \frac{7 !}{4 ! 3 !}+\frac{5 !}{5 ! 0 !} \times \frac{7 !}{3 ! 4 !}$

$=210+175+35=420$