A man X has 7 friends,4 of them are ladies an | vedclass

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(B) Let $L_X = 4, M_X = 3$ be the number of lady and man friends of $X$,and $L_Y = 3, M_Y = 4$ be the number of lady and man friends of $Y$.We need to

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$485$

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(B) Let $L_X = 4, M_X = 3$ be the number of lady and man friends of $X$,and $L_Y = 3, M_Y = 4$ be the number of lady and man friends of $Y$.We need to select $3$ ladies and $3$ men in total,such that $3$ friends are chosen from $X$'s group and $3$ friends are chosen from $Y$'s group.Let $X$ invite $l_1$ ladies and $m_1$ men,and $Y$ invite $l_2$ ladies and $m_2$ men.Constraints: $l_1 + m_1 = 3$,$l_2 + m_2 = 3$,$l_1 + l_2 = 3$,and $m_1 + m_2 = 3$.Possible cases $(l_1, m_1)$ for $X$ and $(l_2, m_2)$ for $Y$ are:Case $1$: $l_1=3, m_1=0$ and $l_2=0, m_2=3$. Ways: $\binom{4}{3}\binom{3}{0} 	imes \binom{3}{0}\binom{4}{3} = 4 	imes 4 = 16$.Case $2$: $l_1=2, m_1=1$ and $l_2=1, m_2=2$. Ways: $\binom{4}{2}\binom{3}{1} 	imes \binom{3}{1}\binom{4}{2} = 18 	imes 18 = 324$.Case $3$: $l_1=1, m_1=2$ and $l_2=2, m_2=1$. Ways: $\binom{4}{1}\binom{3}{2} 	imes \binom{3}{2}\binom{4}{1} = 12 	imes 12 = 144$.Case $4$: $l_1=0, m_1=3$ and $l_2=3, m_2=0$. Ways: $\binom{4}{0}\binom{3}{3} 	imes \binom{3}{3}\binom{4}{0} = 1 	imes 1 = 1$.Total ways $= 16 + 324 + 144 + 1 = 485$.

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