Solution :
5 different mangoes can be distributed by following ways among 3 children such that each gets at least 1 :
Total number of ways : (\(5!\over 3!1!1!2!\) + \(5!\over 2!2!2!\)) \(\times\) 3!
Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children = \(3^7\) (as each fruit has 3 options).
Therefore, Total number of ways = (\(5!\over 3!2!\) + \(5!\over {(2!)}^3\)) \(\times\) 3! \(\times\) \(3^7\)
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