Two families with three members each and

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?

  1. (1) $2 ! 3 ! 4 !$

  2. (2) $(3 !)^{3} \cdot(4 !)$

  3. (3) $(3 !)^{2} \cdot(4 !)$

  4. (4) $3 !(4 !)^{3}$

Correct Option: , 2

Solution:

Number of arrangement

$=(3 ! \times 3 ! \times 4 !) \times 3 !=(3 !)^{3} 4 !$

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