Group A has 7 boys and 3 girls; Group B has 6 boys and 5 girls. A team of 4 boys and 4 girls is to be selected such that exactly 5 members come from Group A and 3 from Group B. The number of ways is
Step-by-Step Solution
Key Concept: Split into cases by the number of boys from Group A (3 or 4), ensuring totals of 4 boys, 4 girls, 5 from A, and 3 from B are all satisfied.
Let (boys from A, girls from A, boys from B, girls from B) satisfy: boys total = 4, girls total = 4, from A = 5, from B = 3.
**Case 1** $(4,1,0,3)$: $\binom{7}{4}\binom{3}{1}\binom{6}{0}\binom{5}{3}=35\cdot3\cdot1\cdot10=1050$.
**Case 2** $(3,2,1,2)$: $\binom{7}{3}\binom{3}{2}\binom{6}{1}\binom{5}{2}=35\cdot3\cdot6\cdot10=6300$.
**Case 3** $(2,3,2,1)$: $\binom{7}{2}\binom{3}{3}\binom{6}{2}\binom{5}{1}=21\cdot1\cdot15\cdot5=1575$.
Total $=1050+6300+1575=8925$.
Correct Answer: 3