How many distinct four-digit integers can one make from the digits $1,3,3,7,7,8$? This is a question from Discrete and combinatorial mathematics book by Ralph Grimaldi .
The question is : How many distinct four-digit integers can one make from the digits $1,3,3,7,7,8$?
In the guidance book, the question is separated into many cases and every case is calculated.  My problem is when we want to calculate the case with one $7$ and two $3$'s, why is the answer $2 \cdot \frac{4!}{2!}$? 
 A: There is $3$ cases:
$1)$ All digits are distinct $= 4!$
$2)$ Exactly two digits are the same = $2 \times 3 \times \frac{4!}{2!}$
$[2$ for choosing $3$ or $7$ to coming twice __ $3$ for choosing $2$ digits from others and the rest for ordering these $4$ digits. Division by $2$ is for repeated counting. You can arrange them in $4!$ ways but you earn same thing when you change the position of same digits. So you count everything twice.$]$
$3)$ Numbers consist of $3,7$ = $\frac{4!}{2! \times 2!}$
$[$ Again $2!$s apear because of double counting.$]$
So the final answer is $24+72+6=102.$
A: For the case in question, if there is exactly one $7$ and two $3$'s, the other digit must be a $1$ or $8$.  We have two ways of choosing this number.  We have $\binom{4}{2}$ ways of choosing the positions of the two $3$'s, which leaves two ways to choose the position of the $7$, and one way to place the remaining digit.
$$2 \cdot \binom{4}{2} \cdot 2 \cdot 1 = 2 \cdot \frac{4!}{2!2!} \cdot 2! = 2 \cdot \frac{4!}{2!}$$ 
The factor of $2!$ in the denominator represents the number of ways we could permute the two $3$'s within a given arrangement without producing an arrangement distinguishable from that arrangement.  
A: When we have same digit or alphabet is in repeating we divide by (number of times)! that thing repeating.
In your case 3 is repeating twice. So we divide by 2!.
