Permutations and Combinations – Selection (Combinations) with Constraints – JEE Main 24 January 2025 Shift 2

Solution:

We need to select a total of 4 Boys and 4 Girls (8 people).
Condition: 5 people from Group A and 3 people from Group B.
We analyze the possible compositions of the 5 people from Group A (Total available: 7 Boys, 3 Girls). The remaining members will be picked from Group B (Total available: 6 Boys, 5 Girls) to satisfy the total requirement of 4 Boys and 4 Girls.

Case I: Group A contributes 2 Boys and 3 Girls
Selection from A: ${}^{7}C_{2} \times {}^{3}C_{3}$.
Remaining needed: 2 Boys and 1 Girl from B.
Selection from B: ${}^{6}C_{2} \times {}^{5}C_{1}$.
Ways $= ({}^{7}C_{2} \times {}^{3}C_{3}) \times ({}^{6}C_{2} \times {}^{5}C_{1}) = (21 \times 1) \times (15 \times 5) = 21 \times 75 = 1575$.

Case II: Group A contributes 3 Boys and 2 Girls
Selection from A: ${}^{7}C_{3} \times {}^{3}C_{2}$.
Remaining needed: 1 Boy and 2 Girls from B.
Selection from B: ${}^{6}C_{1} \times {}^{5}C_{2}$.
Ways $= ({}^{7}C_{3} \times {}^{3}C_{2}) \times ({}^{6}C_{1} \times {}^{5}C_{2}) = (35 \times 3) \times (6 \times 10) = 105 \times 60 = 6300$.

Case III: Group A contributes 4 Boys and 1 Girl
Selection from A: ${}^{7}C_{4} \times {}^{3}C_{1}$.
Remaining needed: 0 Boys and 3 Girls from B.
Selection from B: ${}^{6}C_{0} \times {}^{5}C_{3}$.
Ways $= ({}^{7}C_{4} \times {}^{3}C_{1}) \times ({}^{6}C_{0} \times {}^{5}C_{3}) = (35 \times 3) \times (1 \times 10) = 105 \times 10 = 1050$.

Total ways $= 1575 + 6300 + 1050 = 8925$.

Ans. (3)