Obtain the check digits in this problem

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The check digit for the integer 55382006 is calculated as 7, derived from the weighted sum of its digits modulo 10. For the integer 81372439, the check digit is 5, obtained through a similar calculation. Both check digits are essential for validating the integrity of the respective numbers. The final check digits to be appended are 7 for 55382006 and 5 for 81372439. This method ensures accuracy in numerical verification.
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Homework Statement
Consider the eight-digit bank identification number ## a_{1}a_{2}...a_{8} ##, which is followed by a ninth check digit ## a_{9} ## chosen to satisfy the congruence
## a_{9}\equiv (7a_{1}+3a_{2}+9a_{3}+7a_{4}+3a_{5}+9a_{6}+7a_{7}+3a_{8})\pmod {10} ##.
Obtain the check digits that should be appended to the two numbers ## 55382006 ## and ## 81372439 ##.
Relevant Equations
None.
First, we consider the integer ## 55382006 ##.
Observe that
\begin{align*}
&a_{9}\equiv (5\cdot 7+5\cdot 3+3\cdot 9+8\cdot 7+2\cdot 3+0\cdot 9+0\cdot 7+6\cdot 3)\pmod {10}\\
&\equiv 157\pmod {10}\\
&\equiv 7\pmod {10}.\\
\end{align*}
Thus ## a_{9}=7 ##.
Next, we consider the integer ## 81372439 ##.
Observe that
\begin{align*}
&a_{9}\equiv (8\cdot 7+1\cdot 3+3\cdot 9+7\cdot 7+2\cdot 3+4\cdot 9+3\cdot 7+9\cdot 3)\pmod {10}\\
&\equiv 225\pmod {10}\\
&\equiv 5\pmod {10}.\\
\end{align*}
Thus ## a_{9}=5 ##.
Therefore, the check digits that should be appended to the two numbers ## 55382006 ## and ## 81372439 ## are ## 7 ## and ## 5 ##.
 
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Correct.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

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