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

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(B) $X$ has $4$ ladies and $3$ men. $Y$ has $3$ ladies and $4$ men.We need to select $3$ ladies and $3$ men in total,such that $3$ friends are chosen

Vedclass · Accepted Answer

$485$

Vedclass · Answer

(B) $X$ has $4$ ladies and $3$ men. $Y$ has $3$ ladies and $4$ men.We need to select $3$ ladies and $3$ men in total,such that $3$ friends are chosen from $X$'s group and $3$ friends from $Y$'s group.Let $X$ choose $l_1$ ladies and $m_1$ men,and $Y$ choose $l_2$ ladies and $m_2$ men.Constraints: $l_1 + m_1 = 3$,$l_2 + m_2 = 3$,$l_1 + l_2 = 3$,$m_1 + m_2 = 3$.Possible cases $(l_1, m_1)$ for $X$ and $(l_2, m_2)$ for $Y$:$1. (l_1, m_1) = (3, 0) \implies (l_2, m_2) = (0, 3)$. Ways: $\binom{4}{3}\binom{3}{0} 	imes \binom{3}{0}\binom{4}{3} = 4 	imes 4 = 16$.$2. (l_1, m_1) = (2, 1) \implies (l_2, m_2) = (1, 2)$. Ways: $\binom{4}{2}\binom{3}{1} 	imes \binom{3}{1}\binom{4}{2} = (6 	imes 3) 	imes (3 	imes 6) = 18 	imes 18 = 324$.$3. (l_1, m_1) = (1, 2) \implies (l_2, m_2) = (2, 1)$. Ways: $\binom{4}{1}\binom{3}{2} 	imes \binom{3}{2}\binom{4}{1} = (4 	imes 3) 	imes (3 	imes 4) = 12 	imes 12 = 144$.$4. (l_1, m_1) = (0, 3) \implies (l_2, m_2) = (3, 0)$. 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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