Let $x$ be the number of toffees in the stock initially.
$$\begin{array}{} & \text{Gives} & \text{Left with} \\ 1^{\text{st}} & \frac{x}{2} + 1 & \frac{x}{2}-1 \\ 2^{\text{nd}} & \frac{1}{2}\left( \frac{x}{2} – \right) + 1 & \frac{1}{2}\left(\frac{x}{2} – 1\right) – 1 \\ 3^{\text{rd}} & \frac{1}{2}\left[ \frac{1}{2}\left(\frac{x}{2} – 1\right) – 1 \right] + 1 & \frac{1}{2}\left[ \frac{1}{2}\left(\frac{x}{2} – 1\right) – 1 \right]-1 \\ \vdots & \vdots \quad \vdots \quad \vdots & \vdots \quad \vdots \quad \vdots \\ 5^{\text{th}} & \vdots \quad \vdots \quad \vdots & 0 \end{array}$$
His total stock exhausts after he takes the case of $5$ children.
So, $ \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2}x-1 \right) – 1 \right) – 1 \right) – 1 \right) – 1 = 0$
$ \Rightarrow \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{4}x – \frac{1}{2} – 1 \right) – 1 \right) – 1 \right) – 1 = 0$
$ \Rightarrow \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{8}x – \frac{1}{4} – \frac{1}{2} – 1 \right) – 1 \right) – 1 = 0$
$ \Rightarrow \frac{1}{2} \left(\frac{1}{16}x – \frac{1}{8} – \frac{1}{4} – \frac{1}{2} – 1 \right) – 1= 0$
$ \Rightarrow \frac{1}{32}x – \frac{1}{16} – \frac{1}{8} – \frac{1}{4} – \frac{1}{2} – 1= 0$
$ \Rightarrow \frac{x – 2 – 4 – 8 – 16 – 32}{32} = 0$
$ \Rightarrow x – 62 = 0$
$ \Rightarrow \boxed{x = 62}$
$\therefore 62$ toffees were there in his stock initially.
$\textbf{Second Method:}$
$$\begin{array}{|c|c|c|c|} \hline \text{Toffees} & \text{Children} & \text{Gives} & \text{Left with} \\\hline a\;(\text{Initially}) & 1^{\text{st}} & \frac{a}{2} + 1 & \frac{a}{2}-1 \\ b & 2^{\text{nd}} & \frac{b}{2} + 1 & \frac{b}{2} – 1 \\ c & 3^{\text{rd}} & \frac{c}{2} + 1 & \frac{c}{2} – 1 \\ d & 4^{\text{th}} & \frac{d}{2} + 1 & \frac{d}{2} – 1 \\ e & 5^{\text{th}} & \frac{e}{2} + 1 & \underbrace{\frac{e}{2} – 1}_{0} \\\hline \end{array} $$
Here, $\frac{e}{2} – 1 = 0$
$ \Rightarrow \frac{e}{2} = 1$
$ \Rightarrow \boxed{e=2}$
And, $\frac{d}{2} – 1 = e$
$ \Rightarrow \frac{d}{2} - 1 = 2$
$ \Rightarrow \frac{d}{2} = 3$
$ \Rightarrow \boxed{d = 6}$
Similarly $\frac{c}{2} – 1 = d$
$ \Rightarrow \frac{c}{2} - 1 = 6$
$ \Rightarrow \frac{c}{2} = 7$
$ \Rightarrow \boxed{c = 14}$
And, $\frac{b}{2} – 1 = c$
$ \Rightarrow \frac{b}{2} - 1 = 14$
$ \Rightarrow \frac{b}{2} = 15$
$ \Rightarrow \boxed{b = 30}$
And, $\frac{a}{2} – 1 = b$
$ \Rightarrow \frac{a}{2} - 1 = 30$
$ \Rightarrow \frac{a}{2} = 31$
$ \Rightarrow \boxed{a = 62}$
Correct Answer$: 62$