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option D ?

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The first langauge L1 necessitates that d should follow c and no of c = no of d..And the second langauage L2 says that no of a's and b's should be equal.

In both of the languages , the sequence is a's followed by b's followed by c's and then d's..

But considering the requirement of L1 and L2 , we need that no of a's = no of b's and no of c's  = no of d's ..However no constraint between no of a's and no of c's or no of b's and no of d's..

Hence the language L1 ∩ L2  =  { an bn cm dm  |  m,n >= 0 }

Hence D) is the correct answer

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So, answer-(C) is also right, but aswer-(D) is global case,is it?

will this be possible in c a2b2c3d3 ?

U have to cover all cases right..C) does not cover all strings like aabbcd which should be covered in L1 intersection L2..

Hence C) is wrong..
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