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Permutations and Combinations question

2020 · 6 Sep · Shift 1 · Q38
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Permutations and Combinations question

2020 · 6 Sep · Shift 1 · Q38

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
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.(4!)
  3. C
    3! (4!)3
  4. D
    (3!)2.(4!)
View written solutionFree

Correct answer: B

  1. Interpret the condition

    There are 3 families total:

    • Family 1 has 333 members
    • Family 2 has 333 members
    • Family 3 has 444 members

    The condition “same family members are not separated” means that members of each family must sit together as a single block.

  2. Treat each family as one block

    So we have the following blocks:

    • Block of size 333
    • Block of size 333
    • Block of size 444

    These 333 blocks can be arranged in a row in 3!3!3! ways.

  3. Arrange members within each block

    • Each 3-member family can arrange internally in 3!3!3! ways.
    • Since there are two such families, internal arrangements contribute 3!×3!=(3!)23!\times 3!=(3!)^23!×3!=(3!)2
    • The 4-member family can arrange internally in 4!4!4! ways.

  4. Total number of arrangements

    Hence total ways 3!×3!×3!×4!=(3!)3⋅4!3!\times 3!\times 3!\times 4! = (3!)^3\cdot 4!3!×3!×3!×4!=(3!)3⋅4!

  5. Match with options

    Option B is (3!)3⋅4!(3!)^3\cdot 4!(3!)3⋅4!

    So the correct answer is B.

  6. Compare with stored answer

    Stored correct answer: B

    Our derived answer: B

    They match.

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