Group A consists of 7 boys and 3 girls,while | vedclass

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(C) We need to select $4$ boys and $4$ girls in total,such that $5$ students are from group $A$ and $3$ students are from group $B$.Let $x_1, y_1$ be

Vedclass · Accepted Answer

$8925$

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(C) We need to select $4$ boys and $4$ girls in total,such that $5$ students are from group $A$ and $3$ students are from group $B$.Let $x_1, y_1$ be the number of boys and girls selected from group $A$,and $x_2, y_2$ be the number of boys and girls selected from group $B$.We have $x_1 + y_1 = 5$ and $x_2 + y_2 = 3$,with $x_1 + x_2 = 4$ and $y_1 + y_2 = 4$.Possible cases:Case $I$: $x_1=2, y_1=3$ (from $A$) and $x_2=2, y_2=1$ (from $B$): $\binom{7}{2} \binom{3}{3} 	imes \binom{6}{2} \binom{5}{1} = 21 	imes 1 	imes 15 	imes 5 = 1575$.Case $II$: $x_1=3, y_1=2$ (from $A$) and $x_2=1, y_2=2$ (from $B$): $\binom{7}{3} \binom{3}{2} 	imes \binom{6}{1} \binom{5}{2} = 35 	imes 3 	imes 6 	imes 10 = 6300$.Case $III$: $x_1=4, y_1=1$ (from $A$) and $x_2=0, y_2=3$ (from $B$): $\binom{7}{4} \binom{3}{1} 	imes \binom{6}{0} \binom{5}{3} = 35 	imes 3 	imes 1 	imes 10 = 1050$.Total ways $= 1575 + 6300 + 1050 = 8925$.

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:

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