<s>[INST] Making $6$ digits numbers by the digits $2,3,3,5,5,5,5$ 
How many $6$ digits number we can generate by the digits $2,3,3,5,5,5,5$ ?

To solve this problem I considered three cases because we have $7$ digits and looking for six digit number.

*

*Numbers without $2\Rightarrow\quad\dfrac{6!}{4!2!}=15$


*Numbers without one $3\Rightarrow\quad\dfrac{6!}{4!}=30$


*Numbers without one $5\Rightarrow\quad\dfrac{6!}{3!2!}=60$
So the answer is $15+30+60=105$.

But The book I'm reading from used another method:
Number of $6$ digits permutation from these $7$digits=Number of $7$ digits permutation from these $7$digits= $\dfrac{7!}{2!4!}=105$.
But I don't completely why permutation of  $6$ digits is equal to permutation of all the digits.
 [/INST] \begin{array}{|c|c|c|c|}
\hline\hline
2s&3s&5s&6-\text{digit numbers}\\
\hline
1&1&4&\frac{6!}{4!}=30\\
\hline
1&2&3&\frac{6!}{2!3!}=60\\
\hline
0&2&4&\frac{6!}{2!4!}=15
\end{array}
\begin{array}{|c|c|}
\hline
\hline
\text{Total 6-digit numbers}&30+60+15=105
\end{array}
Your calculations are correct, as shown in the table. Regarding your question as to why the seven-digit permutations work, observe that every six-digit permutation can be changed into a seven-digit permutation by appending the unused digit. There is a one-to-one correspondence between the two.
$$233555\color{magenta}{5}$$
$$253553\color{magenta}{5}$$
$$255553\color{magenta}{3}$$
</s>