There are two solenoids of same length and inductance $L$ but their diameters differ to the extent that one can just fit into the other. They are connected in three different ways in series. (1) They are connected in series but separated by large distance, (2) they are connected in series with one inside the other and senses of the turns coinciding, (3) both are connected in series with one inside the other with senses of the turns opposite as depicted in figures 1,2 and 3 respectively. The total inductance of the solenoids in each of the case 1,2 and 3 are respectively

When two solenoids of inductance $L_0$ are connected in series at large distance and current $i$ is passed through them, the total flux linkage ${\varphi }_{\text{total }}$ is the sum of the flux linkages $L_{0}i$ and $L_{0}i$, i.e.,
${\varphi }_{\text{total }}=L_{0}i+L_{0}i$
If $L$ be the equivalent inductance of the system, then
${\varphi }_{\text{total }}=Li$
$\therefore Li=L_{0}i+L_{0}i$
or $L=2L_0$
When solenoids are connected in series with one inside the other and senses of the turns coinciding, then there will be mutual inductance $L$
between them. In this case the resultant induced emf in the coils is the sum of the emf s $e_1$ and $e_2$ in the respective coils, i.e.,
$e=e_1+e_2$
$=\left(-L_0\frac{di}{dt}\pm L_0\frac{di}{dt}\right)+\left(-L_0\frac{di}{dt}\pm L_0\frac{di}{dt}\right)$
Where $(+)$ sign is for positive coupling and $(-)$ sign for negative coupling.
But, $e=-L\cdot \frac{di}{dt}$
$\therefore -L\frac{di}{dt}=-L_0\frac{di}{dt}-L_0\frac{di}{dt}\pm 2L_0\frac{di}{dt}$
ie, $L=L_0+L_0+2L_0$
$=4L_0$ (for positive coupling)
When solenoids are connected in series with one
inside the other with senses of the turns opposite,
then their is negative coupling.
So, $L=L_0+L_2-2L_0=0$