Example 15 - Chapter 6 Class 11 Permutations and Combinations

Last updated at August 8, 2026 by

Example 15 - In how many ways can 4 red, 3 yellow, 2 green discs - Examples

part 2 - Example 15 - Examples - Serial order wise - Chapter 6 Class 11 Permutations and Combinations

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Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ? Total number of discs = 4 red + 3yellow + 2 green n = 9. Since color are repeating so we use this formula š‘›!/š‘1!š‘2!š‘3! Here n = 9 Out of 9 discs, 4 red, 3 yellow and 2 green & p1 = 4, p2 = 3, p3 = 2 The number of arrangement = š‘›!/š‘1!š‘2!š‘3! = 9!/(4! 3! 2!) = (9 Ɨ 8 Ɨ 7 Ɨ 6 Ɨ 5 Ɨ 4!)/(4! Ɨ (3 Ɨ 2 Ɨ 1)(2 Ɨ 1)) = (9 Ɨ 8 Ɨ 7 Ɨ 6 Ɨ 5)/(3 Ɨ 2 Ɨ 2) = 1260. Hence, total number of arrangements is 1260

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