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?
Solution
F<sub>1</sub> $\to$ 3 members
<br>F<sub>2</sub> $\to$ 3 members
<br>F<sub>3</sub> $\to$ 4 members
<br><br>Total arrangements of three families = 3!
<br><br>Arrangement between members of F<sub>1</sub> family = 3!
<br><br>Arrangement between members of F<sub>2</sub> family = 3!
<br><br>Arrangement between members of F<sub>3</sub> family = 4!
<br><br>$\therefore$ Total numbers of ways can they be seated so that the same family members are not separated
<br><br>= 3! $\times$ 3! $\times$ 3! $\times$ 4!
<br><br>= (3!)<sup>3</sup>.(4!)
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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