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.

Once you have arranged the first six digits, there is only one choice for the seventh digit.  Thus, the number of arranging all seven digits also gives the total number of ways of arranging the first six of those seven digits.
