There are $7$ people including A, B and C. In how many ways can they be arranged such that B is between A and C?
I broke this into cases
$$\_ B\space \_ \space \_ \space \_ \space \_ \space \_ \space $$
$2$ ways to put either A or C in first slot and one of them can be put in remaining 5 slots. Others can be permutated among themselves.
$$\binom{2}{1}\times 5\times 4! \times 2$$
$$\_ \space \_ B\space \_ \space \_ \space \_ \space \_ \space $$
$$2!\times 4\times 4! \times 2$$
$2!$ ways to put either A or C in first slot and one of them can be put in remaining 4 slots. Others can be permutated among themselves.
$$\_ \space \_ \space \_ \space B \_ \space \_ \space \_ \space $$
$$3\times 2 \times 3\times 4! $$
There are $3\times 2 $ ways to put A or C in 3 slots and one of them will seat in remaining 3 slots in $3$ ways. Others can be permutated in $4!$
I also used symmetry in first and second cases. However, my calculations seem to be wrong. Why?
There are $7!$ ways to arrange seven people. By symmetry, in one third of these arrangements, $B$ is between $A$ and $C$, so the number of admissible arrangements is $$\frac{1}{3} \cdot 7!$$
We correct your counts for the individual cases
$B$ is in the second position: There are two ways to choose whether $A$ or $C$ is to the left of $B$ and one way to place the chosen letter in that slot. There are five ways to place the other letter to the right of $B$. The remaining four letters can be arranged in the remaining four positions in $4!$ ways. Hence, there are $$\binom{2}{1}\binom{5}{1}4!$$ such arrangements.
$B$ is in the third position: There are two ways to choose whether $A$ or $C$ is to the left of $B$. Suppose it is $A$. Then there are two ways to choose the position of $A$ and four ways to choose the position of $B$. The remaining four letters can be placed in the remaining four slots in $4!$ ways. Hence, there are $$\binom{2}{1}\binom{2}{1}\binom{4}{1}4!$$ such arrangements.
$B$ is in the fourth position: There are two ways to choose whether $A$ or $C$ is to the left of $B$. Suppose it is $A$. Then there are three ways to choose the position of $A$ and three ways to choose the position of $C$. The remaining four letters can be placed in the remaining four slots in $4!$ ways. Hence, there are $$\binom{2}{1}\binom{3}{1}\binom{3}{1}4!$$ such arrangements.
By symmetry, the number of arrangements with $B$ in the fifth position is equal to the number of arrangements with $B$ in the third position and the number of arrangements with $B$ in the sixth position is equal to the number of arrangements with $B$ in the second position. Hence, the number of admissible arrangements is $$\binom{2}{1}\binom{5}{1}4! + \binom{2}{1}\binom{2}{1}\binom{4}{1}4! + \binom{2}{1}\binom{3}{1}\binom{3}{1}4! + \binom{2}{1}\binom{2}{1}\binom{4}{1}4! + \binom{2}{1}\binom{5}{1}4!$$