solution
this is what I did. total number of ways without any restriction =900. we have to subtract the case when at least one child doesn't get any fruit. let that be C1∪C2∪C3 and we use the principle of mutual inclusion and exclusion to find that. C1∪C2∪C3=C1+C2+C3−(all two intersections)+3 intersections (no one gets a fruit case) =3⋅60−3+1=178 so case where each child gets at least one fruit =900−178=722.
Is this right, someone saying answer =49
