Answer
Yes, it can be done
(8 hexagons and 4 squares).
Work Step by Step
If this can be done, they should make exactly x hexagons and y squares,
where x and y are non-negative integers. (*)
In x hexagons and y squares, there are
x+4y members of the pompom squad, (24 in all),
and
6x+y members of the band, (52 in all),
so we have a system:
$\left\{\begin{array}{lll}
x+4y=24 & \Rightarrow & x=24-4y\\
6x+y=52 & &
\end{array}\right.$
Substituting x with $24-4y$ in the second equation,
$6(24-4y)+y=52$
$144-24y+y=52$
$-23y=52-144\qquad/\div(-23)$
$y=\displaystyle \frac{-92}{-23}=4$
Back substitute into $x=24-4y$
$x=24-4(4)=8$
All conditions (*) are satisfied.
Yes, it can be done
(8 hexagons and 4 squares).
