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 $2 \cdot \frac{4!}{2!}$?
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 $24+72+6=102.$
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.
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!.
