Since at least 3 questions are required to attempt from each part. Therefore a student can attempt (a) 3 questions from part I & 5 questions from part II (b) 4 questions from part I & 4 questions from part II (c) 5 questions from part I & 3 questions from part II In case (a), selection can be made in ${}^{5}C_{3} \times {}^{7}C_{5}$ ways. In case (b), selection can be made in ${}^{5}C_{4} \times {}^{7}C_{4}$ ways. In case (c), selection can be made in ${}^{5}C_{5} \times {}^{7}C_{3}$ ways. ∴ The required no. of ways = ${}^{5}C_{3} \times {}^{7}C_{5} + {}^{5}C_{4} \times {}^{7}C_{4}$ + ${}^{5}C5 \times {}^{7}C_{3}$ = $\frac{5!}{3!2!} \times \frac{7!}{5!2!}$ + $\frac{5!}{4!1!} \times \frac{7!}{4!3!}$ + $\frac{5!}{5!0!} \times \frac{7!}{3!4!}$ = $\frac{5\times4}{2} \times \frac{7\times6}{2}$ + 5 × $\frac{7\times6\times5}{3\times2}$ + 1 × $\frac{7\times6\times5}{3\times2}$ = 210 + 175 + 35 = 420