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📐 Mathematics  ·  Permutations and Combinations  ·  JEE

In how many ways can 4 red, 3 blue, and 2 green balls be arranged in a row?

Answer: 9!/(4!3!2!).

  • A 9!
  • B 9!/(4!3!2!)
  • C 4!3!2!
  • D 24

Correct answer: B. 9!/(4!3!2!)

Explanation: Multinomial permutation: n=9 total, with groups 4,3,2. Arrangements = 9!/(4!3!2!) = 362880/(24×6×2) = 362880/288 = 1260.

StartABCBCACAB6 ordered outcomes (permutations); pairing AB/BA etc gives 3 combinations

Counting tree for selecting 2 items from {A, B, C} without repetition: 3 x 2 = 6 ordered arrangements.

Concept context

Counting, arrangements, and selections

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