Let $f_{1}$ and $f_{2}$ are focal lengths of convex and concave lens respectively, when they are kept in contact co-axially, then focal length $F$ of the combination is given as
$$
\begin{aligned}
\frac{1}{F} &=\frac{1}{f_{1}}+\frac{1}{\left(-f_{2}\right)} \quad\left[\begin{array}{l}
\text { Focal length of concave } \\
\text { lens is negative. }
\end{array}\right] \\
\Rightarrow \quad \frac{1}{F} &=\frac{f_{2}-f_{1}}{f_{1} f_{2}} \\
\Rightarrow \quad F &=\frac{f_{1} f_{2}}{f_{2}-f_{1}}
\end{aligned}
$$
When both lenses are kept at a distance $d$, then their combined focal length $F^{\prime}$ is given as
$$
\begin{aligned}
\frac{1}{F^{\prime}} &=\frac{1}{f_{1}}+\frac{1}{\left(-f_{2}\right)}-\frac{d}{f_{1}\left(-f_{2}\right)} \\
\Rightarrow \quad \frac{1}{F^{\prime}} &=\frac{1}{f_{1}}-\frac{1}{f_{2}}+\frac{d}{f_{1} f_{2}} \\
\Rightarrow \quad \frac{1}{F^{\prime}} &=\frac{f_{2}-f_{1}+d}{f_{1} f_{2}} \\
\Rightarrow \quad F^{\prime} &=\frac{f_{1} f_{2}}{f_{2}-f_{1}+d}
\end{aligned}
$$
From Eqs. (i) and (ii), it is clear that $F^{\prime} < F$ i.e. focal length of combination will decrease.