Question 23

The total area of the figure is $5$ units$^2$.

Let QC $= x$ units.

$\begin{aligned} \text{We want to find } x \text{ such that }\ \ \text{ shaded area} &=2 \frac{1}{2} \text{ units}^2 \\ \\ \therefore\ \text{ area of } \bigtriangleup\!\!\text{PCQ}&= \frac{1}{2} \text{ units}^2 \\ \\ \therefore\ \frac{1}{2} \times 2 \times x &= \frac{1}{2} \\ \\ \therefore\ x &= \frac{1}{2} \end{aligned}$

$\therefore\ $ when Q is the midpoint of [BC], [PQ] divides the figure into equal areas.

The first quartet may be selected in $5^4$ ways, the second in $4^4$ ways, the third in $3^4$ ways, and so on.

Hence the total number of sets of $5$ quartets

$\ =\dfrac{5^4 \times 4^4 \times 3^4 \times 2^4 \times 1^4}{5!} \qquad \{$each set of $5$ quartets can be selected in $5$! different orders$\}$

$\ =\dfrac{(5!)^4}{5!}$

$\ =(5!)^3$

$\ =1\,728\,000$