<s>[INST] Numbers formed from all the digits $1113333344455678$ so that $6$ and $8$ are on dofferent sides from $7$ 
How many numbers formed from all the digits $1113333344455678$ are there if the digits $6$ and $8$ have to appear from the different sides of the digit $7$?

My thoughts:
I thought I should first find how many numbers are formed from all the $14$ digits $11133333444557,$ which is $$\binom{14}3\binom{11}5\binom63\binom32=\frac{14!}{3!5!3!2!},$$and, then, analyze the possibilies given a position of the digit $7,$ but then, I lose the track and I only tell for sure $6$ and $8$ can be first and last digits in every situation, so there are strictly more numbers than $2\frac{14!}{3!5!3!2!}=\frac{14!}{(3!)^25!}.$
Then, I thought, I should first consider the position $p, 1<p<16$ among $16$ overall places for $7$ and choose the positions in $p-1$ and $16-p$ ways for $6$ and $8$ and multiply the result by $2$ for each $p,$ but this didn't lead me anywhere.
How should I approach this problem?
 [/INST] One of the approaches is to see that once three  positions are decided for $6, 7$ and $8$, you can arrange them within, in $3! = 6$ ways but here instead of $6$, there are only $2$ ways to arrange them as $7$ must be in between $6$ and $8$.
So we first find all possible arrangements of $1113333344455678$ and then divide by $3$ to get all favorable arrangements.
So the answer is $ ~ \displaystyle \frac{1}{3} \cdot \frac{16!}{3! \cdot 5! \cdot 3! \cdot 2!}$

Also sharing a longer approach - first arrange $13$ numbers other than $6, 7$ and $8$. Then there are $14$ places between them (including the ends). We can choose three places, two places or one place out of $14$ places for $6, 7$ and $8$. That leads to,
$ \displaystyle \frac{13!}{3! \cdot 5! \cdot 3! \cdot 2!} \left[2 \cdot {14 \choose 1} + 4 \cdot {14 \choose 2} + 2 \cdot {14 \choose 3}\right]$
Note that when we choose two places, if the first place from the left has two numbers then it must have $67$ or $87$ and if the second place has two numbers then it must have $76$ or $78$. That is why we multiply second term by $4$.
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