How many different 9-0 letter code words can be made using the symbols $\%,\%,\%,\%,\And ,\And ,\And ,+,+$ ?

Hint: We have to find the total number of distinguishable ways in which the given symbols of percentage, ampersand and plus sign can be arranged to form 9-0 letter codes. This implies that we have to find the total number of permutations for the given symbols. Thus, we shall apply the formula of finding permutations and calculate the factorial of the total number of given symbols. Further we shall divide it by the factorial of the number of occurrences of each symbol in the given set of symbols.

Complete step by step solution:
Given the symbols, $\%,\%,\%,\%,\And ,\And ,\And ,+,+$.
The formula of permutations for finding the total permutations, $p$ of the word, numbers is given as
$p=\dfrac{n!}{{{m}_{a}}!.{{m}_{b}}!.....{{m}_{z}}!}$
Where,
$n=$total number of letters in the given word
${{m}_{a}},{{m}_{b}},.....,{{m}_{z}}=$ number of occurrences of the letters $a,b,.....,z$ in the given word
The total number of letters in this word are 9, thus, $n=9$.
Here, ${{m}_{\%}},{{m}_{\And }},$and ${{m}_{+}}$ are the total number of occurrences of the symbols $\%,\And $ and $+$
respectively.
We see that ${{m}_{\%}}=5$, ${{m}_{\And }}=3$, and ${{m}_{+}}=2$,
Thus, we get number of permutations as
$p=\dfrac{n!}{{{m}_{\%}}!.{{m}_{\And }}!.{{m}_{+}}!}$
$\Rightarrow p=\dfrac{9!}{5!.3!.2!}$
We know that $9!=362880,\text{ }5!=120,\text{ }3!=6$ and $2!=2$. Substituting these values, we get
$\Rightarrow p=\dfrac{362880}{120.6.2}$
$\Rightarrow p=\dfrac{362880}{1440}$
$\Rightarrow p=252$
Therefore, the total number of different 9-0 letter code words that can be made using the symbols $\%,\%,\%,\%,\And ,\And ,\And ,+,+$ is 252.

Note: One possible mistake we can make while solving these problems related to these permutations of a such 9-0 letter codes is that we can make a mistake while counting the number of repetitive symbols in the given particular set of symbols. Thus, this counting of symbols must be done carefully.