A group of $630$ children seated in rows for a group photo session. Each row contains three less children than the row in front of it. Which one of the following number of rows is not possible?

Let the number of rows be $n$. Let the number of children in the front row be $a$.
Since each successive row contains three less children than the row in front of it, the number of children in each row forms an Arithmetic Progression with common difference $d=-3$.

The total number of children is given by the sum of an AP:

$$S_n=\frac{n}{2}[2a+(n-1)d]=630$$

$$\frac{n}{2}[2a+(n-1)(-3)]=630$$

$$n[2a-3n+3]=1260$$

$$2an-3n^2+3n=1260$$

$$2an=1260+3n^2-3n$$

$$a=\frac{1260+3n^2-3n}{2n}$$

For a configuration to be valid, $a$ must be a positive integer, and the last row must have at least $1$ child. The last row has $a-3(n-1)$ children, so $a-3n+3\ge 1⟹2a-6n+6\ge 2$.

Let's test the options for $n$:

1. If $n=3$:

$$a=\frac{1260+3(3{)}^2-3(3)}{2(3)}=\frac{1260+27-9}{6}=\frac{1278}{6}=213$$

This is a positive integer. The last row has $213-3(2)=207>0$. This is possible.
2. If $n=4$:

$$a=\frac{1260+3(4{)}^2-3(4)}{2(4)}=\frac{1260+48-12}{8}=\frac{1296}{8}=162$$

This is a positive integer. The last row has $162-3(3)=153>0$. This is possible.
3. If $n=5$:

$$a=\frac{1260+3(5{)}^2-3(5)}{2(5)}=\frac{1260+75-15}{10}=\frac{1320}{10}=132$$

This is a positive integer. The last row has $132-3(4)=120>0$. This is possible.
4. If $n=6$:

$$a=\frac{1260+3(6{)}^2-3(6)}{2(6)}=\frac{1260+108-18}{12}=\frac{1350}{12}=112.5$$

Since $a=112.5$ is not an integer, it is impossible to have a fractional number of children in the front row. Therefore, $6$ rows is not a possible arrangement.