Suppose A1,A2,....A30 are thirty sets each with five elements and B1,B2,....Bn are n sets each with three elements such that ∪i=130Ai∪j=1nBj=S . If each element of S belongs to exactly ten of the Ais and exactly 9 of the Bjs , then the value of n is ___

S=∪J=1nAi⇒n(S)=110(5×30)=15 Again S=∪J=1nBj⇒n(S)=19(3×n)=n3 Thus n3=15⇒n=45

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Suppose $A_{1}, A_{2}, .... A_{30}$ are thirty sets each with five elements and $B_{1}, B_{2}, .... B_{n}$ are n sets each with three elements such that $\cup_{i = 1}^{30} A_{i} \cup_{j = 1}^{n} B_{j} = S$ . If each element of S belongs to exactly ten of the $A_{i} s$ and exactly 9 of the $B_{j} s$ , then the value of n is ___

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answer is 45.

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Detailed Solution

$S = \cup_{J = 1}^{n} A_{i} \Rightarrow n (S) = \frac{1}{10} (5 \times 30) = 15 A g a i n S = \cup_{J = 1}^{n} B_{j} \Rightarrow n (S) = \frac{1}{9} (3 \times n) = \frac{n}{3} T h u s \frac{n}{3} = 15 \Rightarrow n = 45$

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