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. ${\left(3!\right)}^3\times 4!$
• C. $3!\times {\left(4!\right)}^3$
• D. ${\left(3!\right)}^2\times 4!$

Answer 1

The correct option is B

${\left(3!\right)}^3\times 4!$

Explanation for the correct answer:

Assume that the all the members of the families are together

Consider the the three families as three separate entities

The three families can be arranged in $3!$ ways

Now the three members in Family $1$ can be arranged in $3!$ ways

Similarly, the three members in Family $2$ can be arranged in $3!$ ways

Similarly, the four members in Family $3$ can be arranged in $4!$ ways

Hence, the total number of favorable arrangements are given as

Number of favorable arrangements $=$Arrangements of families $\times$Arrangements of family members

$\Rightarrow$ Number of favorable arrangements $=$$3!\times 3!\times 3!\times 4!$

$\Rightarrow$ Number of favorable arrangements $=$${\left(3!\right)}^3\times 4!$

Hence, option $\left(B\right)$ is the correct answer