a1, a2, a3 ... a30-each of these 30 sets has 5 elements. b1, b2, ... bn-each of these n sets has 3 elements. union of a1, a2 ... a30=union of b1, b2 ... bn=S.

if each elements of S is in 10 a sets and 9 bsets. then n=?

the number of elements in the union of the A sets is:5 (30) - rAwhere r is the number of repeats. Likewise the number of elements in the B sets is:3n-rB

Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have:150-9x=3n-8x⟺150-x=3n⟺50-x3=n

Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that:150 / 10=15is the number of elements in the the A's without repeats counted (same for the Bs as well). So now we have:50-15 / 3=n⟺n=45