(i) When committee consists of exactly 3 girls. Then remaining will be 4 boys in the committee. So the required no. of ways = $\,{}^{9}C_{4} \times \,{}^{4}C_{3}$ = $\frac{9!}{4!5!} \times \frac{4!}{3!1!}$ = $\frac{9\times 8\times 7\times 6}{6}$ = 504 (ii) When committee consists of atleast 3 girls i.e., it may be 3 girls or 4 girls. Then there are two cases. One when there are 3 girls in a committee, then the boys will be 4 & second when there are 4 girls in a committee, then the boys will be 3. So, No. of ways in first case = $\,{}^{4}C_{3} \times \,{}^{9}C_{4}$, No. of ways in second case = ${}^{4}C_{4} \times {}^{9}C_{3}$ \ Required no. of ways = $\,{}^{4}C_{3} \times \,{}^{9}C_{4} + \,{}^{4}C_{4} \times \,{}^{9}C_{3}$ (iii) When atmost 3 girls are there in committee. Total no. of people = 13, No. of people in committee = 7. No. of ways of 7 people out of 13 people = $\,{}^{13}C_{7}$ = 1716 When 4 girls are in committee, then no. of ways = $\,{}^{4}C_{4} \times \,{}^{9}C_{3}$ = 84. So the required no. of ways = 1716 – 84 = 1632