In how many can three white balls, four black balls and five red balls be arranged.
Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM
In totality we have 3 + 4 + 5 = 12 balls of which 3 are of one kind, 4 are of second kind and 5 are of third kind.
Hence, the distinct permutations of the balls
Answered by | 4th Jun, 2014, 03:23: PM
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