Question

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?

• A. $2!3!4!$
• B. $(3!{)}^3\cdot (4!)$
• C. $3!(4!{)}^3$
• D. $(3!{)}^2\cdot (4!)$

Answer 1

The correct option is B $(3!{)}^3\cdot (4!)$

Family $1=3$ members
Family $2=3$ members
Family $3=4$ members
Total arrangements of $3$ families$=3!$
Arrangement between members of Family $1$ $=3!$
Arrangement between members of Family $2$ $=3!$
Arrangement between members of Family $3$ $=4!$

$\therefore$ Total numbers of ways can they be seated so that the same family members are not separated
$=3!\times 3!\times 3!\times 4!$
$=(3!{)}^3\cdot (4!)$