Group A consists of $7$ boys and $3$ girls, while Group B consists of $6$ boys and $5$ girls. The number of ways $4$ boys and $4$ girls can be invited for a picnic if $5$ of them must be from Group A and the remaining $3$ from Group B, is equal to:
Step-by-Step Solution
Key Concept: Let $a$ = boys chosen from A. Then girls$_A=5-a$, boys$_B=4-a$, girls$_B=a-1$. Feasibility constraints on each give $a\in\{2,3,4\}$. Sum the three cases.
Let $a=$ boys chosen from A. Then
(boys$_A$, girls$_A$, boys$_B$, girls$_B$) = $(a,\,5-a,\,4-a,\,a-1).$
Feasibility: $0\le 5-a\le 3$ and $a-1\ge 0$ give $a\in\{2,3,4\}.$
$a=2$: $\binom{7}{2}\binom{3}{3}\binom{6}{2}\binom{5}{1}=21\cdot 1\cdot 15\cdot 5=1575.$
$a=3$: $\binom{7}{3}\binom{3}{2}\binom{6}{1}\binom{5}{2}=35\cdot 3\cdot 6\cdot 10=6300.$
$a=4$: $\binom{7}{4}\binom{3}{1}\binom{6}{0}\binom{5}{3}=35\cdot 3\cdot 1\cdot 10=1050.$
Total: $1575+6300+1050=8925.$
Correct Answer: 3