Csdrhrt

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

$$_ Bspace _ space _ space _ space _ space _ space $$

$2$ 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.

$$binom{2}{1}times 5times 4! times 2$$

$$_ space _ Bspace _ space _ space _ space _ space $$

$$2!times 4times 4! times 2$$

$2!$ 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.

$$_ space _ space _ space B _ space _ space _ space $$

$$3times 2 times 3times 4! $$

There are $3times 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
$$binom{2}{1}binom{5}{1}4! + binom{2}{1}binom{2}{1}binom{4}{1}4! + binom{2}{1}binom{3}{1}binom{3}{1}4! + binom{2}{1}binom{2}{1}binom{4}{1}4! + binom{2}{1}binom{5}{1}4!$$

First choose 3 places where those 3 will be, that you can do on ${7choose 3}$ ways.

On the middle of choosen places put $B$ and on other two put $A$ and $C$, that you can do on 2 way. Then permute all others on 4! way on remaining places.

So the answer is $${7choose 3}cdot 2cdot 4!$$