Question

Suppose A1,A2,...,A30 ARE 30 SETS EACH WITH FIVE ELEMENTS AND BI, B2, ... Bn are n sets each with three elements. let A1 uA2 U ...UA30 = B1 U B2 U ...U Bn =S. Assume that each element of S belongs to exactly ten of the A's and to exactly nine of the B's. Find n

Answer 1

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

Hope it helps....

Answer 2

Hello,

Here's your answer.

I simplified it as much as i could.

Hope it helps.