I am missing something in this and I need to discuss it . Step 1) We can arrange numbers [4 ,7] with $4!$ , different ways . Step 2) There are 4 dinstinct cases:
- a) We are going to put '2' , '1' and '3' in 3 of the 5 available spaces between numbers [4,8] _N_N_N_N_. There are $C(5,3)=10$ ways to do that
b) We are going to put '21' and '3' or '2' and '13' there are $C(5,2)=10$ to do this for each case so $2*C(5,2)=20$ in total quick fix : I only counted the cases that 2 1 3 come in that order. So to fix this we need to change the positions of 2 and 3 so we have $2!*840=1680$
c) we are going to put ' 213' and there are 5 ways to do this
Sum : $4!*(10 + 20 + 5) = 24 * 35 = 840$ ways But this not the correct answer. Any ideas?
You are counting the cases where $2$ appears first, then $1$, then $3$. However, when the problem asks you to compute permtuations where "$1$ appears between $2$ and $3$," they mean that it is also valid for $3$ to appear first, then $1$, then $2$. For example, in $3415672$, it is valid to say that $1$ is between $2$ and $3$.
It does not take much more work to include these cases, can you see the quick fix?
By they way, there is a quicker way to do this problem. In each of the $7!$ permutations of $\{1,\dots,7\}$, the number of permutations where $1$ is between $2$ and $3$ equals the number of permtuations where $2$ is between $1$ and $3$, which is in turn equal to the number of permutaitons where $3$ is between $1$ and $2$. This is because switching around the names of the numbers does not change the structure of the problem. Since all three of these cases are equal, the number of each must be $7!/3$.