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CA Foundation · Quantitative Aptitude · Permutations and Combinations

A box contains 5 red, 4 blue and 3 green distinct balls. In how many ways can 4 balls be selected so that at least one ball of each colour is included?

The answer is 270. Since 4 balls cover 3 colours, exactly one colour is taken twice. Red twice gives 120 ways, blue twice gives 90 and green twice gives 60. Adding these three disjoint cases gives 270 selections.

  1. A270Correct
  2. B225
  3. C180
  4. D300

Explanation

With 4 balls and 3 colours, one colour appears twice. Red twice: 5C2×4×3 = 10×12 = 120. Blue twice: 4C2×5×3 = 6×15 = 90. Green twice: 3C2×5×4 = 3×20 = 60. Total = 120+90+60 = 270.

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