JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Permutations and Combinations

Need 4 men and 4 women total, drawn as 4 from Group A (4M, 5W) and 4 from Group B (5M, 4W). Splitting by how many men come from A ($m_A$) and the rest from B ($4-m_A$ men from B, contributing $4-(4-m_A)=m_A$... working directly by cases of (men from A, women from A):

• (4M,0W from A; 0M,4W from B): $\left(\genfrac{}{}{0ex}{}{4}{4}\right)\left(\genfrac{}{}{0ex}{}{4}{4}\right)=1$
• (3M,1W from A; 1M,3W from B): $\left(\genfrac{}{}{0ex}{}{4}{3}\right)\left(\genfrac{}{}{0ex}{}{5}{1}\right)\times \left(\genfrac{}{}{0ex}{}{5}{1}\right)\left(\genfrac{}{}{0ex}{}{4}{3}\right)=400$
• (2M,2W from A; 2M,2W from B): $\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{5}{2}\right)\times \left(\genfrac{}{}{0ex}{}{5}{2}\right)\left(\genfrac{}{}{0ex}{}{4}{2}\right)=3600$
• (1M,3W from A; 3M,1W from B): $\left(\genfrac{}{}{0ex}{}{4}{1}\right)\left(\genfrac{}{}{0ex}{}{5}{3}\right)\times \left(\genfrac{}{}{0ex}{}{5}{3}\right)\left(\genfrac{}{}{0ex}{}{4}{1}\right)=1600$
• (0M,4W from A; 4M,0W from B): $\left(\genfrac{}{}{0ex}{}{5}{4}\right)\times \left(\genfrac{}{}{0ex}{}{5}{4}\right)=25$

Total $=1+400+3600+1600+25=5626$.