Two families with three members each and one | vedclass

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(B) Let the three families be $F_1, F_2,$ and $F_3$. The number of members in these families are $3, 3,$ and $4$ respectively.Since the members of the

Vedclass · Accepted Answer

$(3!)^2 \cdot 4! \cdot 3!$

Vedclass · Answer

(B) Let the three families be $F_1, F_2,$ and $F_3$. The number of members in these families are $3, 3,$ and $4$ respectively.Since the members of the same family must stay together,we can treat each family as a single unit.There are $3$ such units (families) which can be arranged among themselves in $3!$ ways.Within each family,the members can be arranged among themselves:- Family $1$ ($3$ members) can be arranged in $3!$ ways.- Family $2$ ($3$ members) can be arranged in $3!$ ways.- Family $3$ ($4$ members) can be arranged in $4!$ ways.Therefore,the total number of ways is $3! 	imes (3! 	imes 3! 	imes 4!) = (3!)^2 	imes 3! 	imes 4! = 6 	imes (3!)^2 	imes 4!$.However,looking at the provided options,the expression $(3!)^2 \cdot 3! \cdot 4!$ simplifies to $6 \cdot (3!)^2 \cdot 4!$. Given the structure of the options,the correct arrangement is $3! \cdot 3! \cdot 3! \cdot 4!$.

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the members of the same family are not separated?

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