# Exponential generating function for making password.

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.

• 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. – P. Quinton Apr 6 '20 at 13:30

• 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?]