Identifying origin in convolution table
Given a signal $x[n]$ with the nonzero range of $N_1 \leq n \leq N_2$ and the signal $h[n]$ with the nonzero range $M_1 \leq n \leq M_2$, then the nonzero range of their comvolution $y[n] = x[n] \star h[n]$ is $N_1+M_1 \leq n \leq N_2 + M_2$.

Applying this to your case yields the following:
For the signal $x[n]$ with $n=0$ at third position the nonzero range is $-2 \leq n \leq 2$ and for $h[n]$ with $n=0$ at the second position the nonzero range is $-1 \leq n \leq 2$ which therefore yields the nonzero range of their convolution $y[n]$ is: 
$$ -3 \leq n \leq 4 $$

which means that the $n=0$ for $y[n]$ is at the fourth position.