Number of String present in language

Question

The number of strings present of length 10 in language  $L=\left \{ a^{2n+1}.b^{2m+1}|n\geq 0,m\geq 0 \right \} $ are_________

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My explanation: I think here minimum string ab , and $8$ gap is present. So, like counting we can solve it. That means  these $8$ place we can fill with either $a$ or $b$

Where is wrong in it?

Comments

i hope there is no need of waiting for srestha mam, answer !

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yes , I understood :)

but can we not think about general case?

like, there will be $8$ space and we can fill them individually?

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My explanation: I think here minimum string ab , and 8 gap is present. So, like counting we can solve it. That means  these 8 place we can fill with either a or b.

Where is wrong in it?

then a  abababab b also possible, right ?

 

but can we not think about general case?

total 10 blanks ===> first should be fill with a and last should be fill with b. ==> remaining 8 blanks

make 2 blanks as one box, due to preserve the condition 2n+1 or 2m+1 ( Note that already 1 is takes place. )

So, Now how many Boxes ? 4 boxes

divide these boxes as two partitions, WHY? due to one partition will allocate a's and other one allocate to b's

i) 0 boxes to a and 4 boxes to b

ii) 1 boxes to a and 3 boxes to b

iii) 2 boxes to a and 2 boxes to b

iv) 3 boxes to a and 1 boxes to b

v) 4 boxes to a and 0 boxes to b

 

each case will lead to exactly one choice of a's and b's into the 8 blanks

ex:-

    ii) 1 boxes to a and 3 boxes to b ===> $a \;\color{red}{aa} \; \color{green}{bb bb bb }\; b$

∴ 5 cases lead to 5 strings of length 10.

Answer 1

only odd powers of a and b are possible.

So possible strings of length 10 will be $a^1b^9 , a^3b^7 ,a^5b^5 , a^7b^3 , a^9b^1$

Only 5 strings possible