If elements are not repeated then number of elements inA1⋃A2⋃A3⋃….⋃A30 = 30*5=150
But, each element of S is in 10 of Ai
∴ s = \(\frac{150}{10}\) = 15.
If elements are not repeated then number of elements in B1⋃B2⋃B3⋃….⋃Bn=3*n
But, each element of S is in 9 of B1
∴ s = \(\frac{3\times{n}}{9}\)
∴ s = \(\frac{3\times{n}}{9}\) = 15
∴ 3 × n =15 × 9
∴ n = 45