No of ways in which 2 alike apples and 4 alike mangoes can be distributed in 3 children so that each child receives any number of fruits.
Task A: Distributing Apples To distribute apples I imagined 2 partitions between 2 alike apples such that apples before partition 1 goes to Student1 and apples after partition 2 goes to Student B also and those in middle goes to Student 3. So number of ways of arranging 2 partitions and 2 apples is $4!/2!×2!$.
Task B: Distributing Mangoes Following a similar analysis number of ways is $6!/4!×2!$. So by multiplication theorem Total ways of completing both tasks is $6!/2!×2!×2!$. Am I right?
Yes, you are correct. Distributing $n$ alike fruits among $k$ children: we count the number of nonnegative integer solutions of $$x_1+x_2+\dots +x_k=n\implies \binom{n+k-1}{k-1}.$$ So, in your case, by the multiplication theorem, the total number of ways is $$\binom{4}{2}\cdot \binom{6}{2}=6\cdot 15=90.$$