Exponential generating function for making password.

combinatoricsdiscrete mathematicsgenerating-functions

There are special symbols $\{$!,@,#$\} $, three alphabet {a,b,c} and three numbers $\{1,2,3\}$

To make the $n$-character password, the following rules are required.

(A) Special symbol should be used only one time.

(B) Alphabets should be used even times including the $0$

(C) The numbers should be used odd times

Find the exponential generating function for making $n$-character password.

e.g.) when $n=4$, $!123$ can be $4$-character password that satisfying the above rules.

Let's consider the exponential generating function, $f(x)$.

Then $f(x) = 3x(1+ {1\over2!}(3x)^2 + {1\over4!}(3x)^4+…)(3x+ {1\over3!}(3x)^3+{1\over5!}(5x)^5+…)$

$\therefore f(x) = 3x({e^{3x}+e^{-3x}\over2 })({e^{3x}-e^{-3x}\over2 })$

But In the solution sheet that my colleagues made said

$f(x) = 3x({e^{x}+e^{-x}\over2 })^3({e^{x}-e^{-x}\over2 })^3$

I don't know why the answer should be like that. At least I believe his answer is not true.(It looks like mine is correct for me.)

What do you think about that? Any help or solution would be appreciated.

Best Answer

It depends how you interpret rules (B) and (C).

  • If (B) means that the total number of alphabet letters must be even, so aabc123 is legal, then your exponential generating function is correct.

  • If (B) means that each alphabet letter must appear an even number of times, so that aabc123 is illegal, then your colleagues are correct. [Can you see why?]

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