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If L1 = { a^n | n ≥ 0 } and L2 = { b^n | n ≥ 0 }, Consider

then L1 . L2 wil be

a) (ab)^n

b) a^n b^n

c) b^n a^n

d)b^m a^n

e)  { a^m b^n | m ≥ 0, n ≥ 0 }

why answer is d why not b ????
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5 Answers

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'n' of  L1 is different from the 'n' of L2 (i.e these two 'n' are not equal), So due to this difference L1.L2 = {a^mb^n :m>=0,n>=0} will be the answer. 

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L1 and L2 are both regular languages, therefore concatenation of both i.e L1.L2 is also regular language. And this language is "Number of a's followed by number of b's".
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no option is correct:- D must be like this

D. ambn and m,n >=0   which also represent a*b*...

L1={null,a,aa,aaa,......}=a*

L2={null,b,bb,bbb,......}=b*

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