We have - $F$, $M$, $S$, $D_1$, and $D_2$

Let us make the cases:

Case I:
$D_1$ and $F$ in front with $D_2$, $S$ and $M$ at the back
Possible Arrangements $= 3! = 6$

Case II:
$D_1$ and $M$ in front with $D_2$, $S$ and $F$ at the back
Possible Arrangements $= 3! = 6$

Case III:
$D_2$ and $F$ in front with $D_1$, $S$ and $M$ at the back
Possible Arrangements $= 3! = 6$

Case IV:
$D_2$ and $M$ in front with $D_2$, $S$ and $F$ at the back
Possible Arrangements $= 3! = 6$

Case V:
$S$ and $F$ in front with $D_1$, $D_2$ and $M$ at the back
Possible Arrangements $= 2$

Case VI:
$S$ and $M$ in front with $D_1$, $D_2$ and $F$ at the back
Possible Arrangements $= 2$

Case VII:
$M$ and $F$(driving) in front with $D_1$, $D_2$ and $S$ at the back
Possible Arrangements $= 2$

Case VIII:
$F$ and $M$(driving) in front with $D_1$, $D_2$ and $S$ at the back
Possible Arrangements $= 2$

Total possibilities $= 6 + 6 + 6 + 6 + 2 + 2 + 2 + 2 = 32$

Hence, option B