in Set Theory & Algebra
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please explain!!

in Set Theory & Algebra
565 views

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min value of n should be 7

Is't correct ??
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no the solution is 63
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this is the solution they have given!!

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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
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