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In how many ways can 12 different( not same) coloured balls be distributed among 3 boys so that each gets atleast 1 ball?  ans:- 519156

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One way : S(12, 3) * 3!

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Other way: Principle of inclusion-exclusion

312 - [ $\binom{3}{1}$ * 212 - $\binom{3}{2}$ * 112]

= 519156