The number of ways of giving 20 distinct oran | vedclass

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(A) To distribute $n$ distinct items into $k$ distinct groups such that no group is empty,we use the Principle of Inclusion-Exclusion.Here,$n = 20$ an

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$3^{20} - 3 	imes 2^{20} + 3$

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(A) To distribute $n$ distinct items into $k$ distinct groups such that no group is empty,we use the Principle of Inclusion-Exclusion.Here,$n = 20$ and $k = 3$.The total number of ways to distribute $20$ distinct oranges to $3$ children without any restriction is $3^{20}$.Let $S$ be the set of all distributions,and $A_i$ be the property that child $i$ receives no orange.We want to find the number of ways where no child receives zero oranges,which is given by $|S| - |A_1 \cup A_2 \cup A_3|$.By the Principle of Inclusion-Exclusion,this is $3^{20} - \binom{3}{1} 2^{20} + \binom{3}{2} 1^{20} - \binom{3}{3} 0^{20}$.$= 3^{20} - 3 	imes 2^{20} + 3 	imes 1 - 0 = 3^{20} - 3 	imes 2^{20} + 3$.

The number of ways of giving $20$ distinct oranges to $3$ children such that each child gets at least one orange is $............$.

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