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JEE Mains · Maths · STD 11 - 6. permutation and combination

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

Answer & Solution

Correct Answer

(B) \((3 !)^{3} \cdot(4 !)\)

Step-by-step Solution

Detailed explanation

Total numbers in three familes \(=3+3+4=10\) so total arrangement \(=10 !\) \( \begin{array}{|c|c||c|}\hline \text { Family 1 } & \text { Family 2 } & \text { Family 3 } \\3 & 3 & 4 \\\hline\end{array}\) Favourable cases…
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