Wronsonia8g

2022-07-03

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
$\left(\genfrac{}{}{0}{}{14}{3}\right)\left(\genfrac{}{}{0}{}{11}{5}\right)\left(\genfrac{}{}{0}{}{6}{3}\right)\left(\genfrac{}{}{0}{}{3}{2}\right)=\frac{14!}{3!5!3!2!},$,
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 $2\frac{14!}{3!5!3!2!}=\frac{14!}{\left(3!{\right)}^{2}5!}.$.
Then, I thought, I should first consider the position $p,1 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.

Amir Beck

Expert

Step 1
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.
Step 2
So we first find all possible arrangements of 1113333344455678 and then divide by 3 to get all favorable arrangements.