Ohh yeah I did it in a wrong way
I explain you properly now
A1 A2 A3 A4 A5 .... A45 --- > total 45 sets
1 sets contains ---- > 7 elements
45 sets contains ------> 45 x 7 elements total
Now we did union -- >A1 $\cup$ A2 $\cup$ A3 ....... $\cup$ A45
S = { A1 $\cup$ A2 $\cup$ A3 ....... $\cup$ A45 }
No each elements in set S is occurs 15 times in Ai 's
now , union of 315 elements and each elements in the union occurs 15 times in Ai's
therefore number of distinct element present = 315 /15 = 21
Now we get the no of element we get after did union = 21
Now similarly ,
B1 B2 B3 B4 B5 .... Bn --- > total "n" sets
1 sets contains ---- > 4 elements
n sets contains ------> nx 4 elements total
now , union of 4n elements and each elements in the union occurs 12 times in Bi's
therefore number of distinct element present = 4n /12 = 21
n = 63 Ans