Two families with three members each and one | vedclass

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(B) There are three families: two families with $3$ members each and one family with $4$ members.To ensure that members of the same family are not sep

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) There are three families: two families with $3$ members each and one family with $4$ members.To ensure that members of the same family are not separated,we treat each family as a single unit.There are $3$ such units (families) to be arranged among themselves,which can be done in $3!$ ways.Within the first family of $3$ members,they can be arranged in $3!$ ways.Within the second family of $3$ members,they can be arranged in $3!$ ways.Within the third family of $4$ members,they can be arranged in $4!$ ways.Therefore,the total number of ways to seat them such that family members stay together is $3! \cdot 3! \cdot 3! \cdot 4! = (3!)^3 \cdot 4!$.

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 same family members are not separated?

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