A family consisting of a mother,father,and th | vedclass

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Question

(A) The family consists of $10$ people: $1$ mother,$1$ father,$4$ boys,and $4$ girls. Total males = $1$ (father) + $4$ (boys) = $5$. Total females = $

Vedclass · Accepted Answer

$576$

Vedclass · Answer

(A) The family consists of $10$ people: $1$ mother,$1$ father,$4$ boys,and $4$ girls. Total males = $1$ (father) + $4$ (boys) = $5$. Total females = $1$ (mother) + $4$ (girls) = $5$. Since the males and females must alternate,we arrange the $5$ males around the circular table in $(5-1)! = 4! = 24$ ways. This creates $5$ spaces between the males. We need to place the $5$ females in these $5$ spaces. However,the mother and father must sit together. Let the father be at position $F_1$. The mother must be in one of the two spaces adjacent to the father. There are $2$ choices for the mother's position relative to the father. The remaining $4$ females can be arranged in the remaining $4$ spaces in $4! = 24$ ways. Total ways = $24 	imes 2 	imes 24 = 576$.

$A$ family consisting of a mother,father,and their $8$ children ($4$ boys and $4$ girls) are to be seated at a round table in a party. How many ways can this be done if the mother and father sit together and the males and females alternate?

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