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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.

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  • $\begingroup$ You can check by developing some coefficients, for instance for $n=2$, you get coefficient $9$ and he gets $0$, which one is right for this example ? From what I understood none of you is because you have $sa$ and $as$ where $s$ is a symbol and $a$ a number, which is $18$ possibilities in total. $\endgroup$ – P. Quinton Apr 6 '20 at 13:30
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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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  • $\begingroup$ Ah! Yes. I understood. Thanks for your comment! :) $\endgroup$ – se-hyuck yang Apr 6 '20 at 22:47

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