Problem: (From Hamilton Maths Olympiad 2015)
Some boys and girls are standing in a row, in some order, about to be photographed. All of them are facing the photographer. Each girl counts the number of boys to her left, and each boy counts the number of girls to his right.
Let the sum of the numbers counted by the girls be $G$, and the sum of the numbers counted by the boys be $B$.
Prove that $G = B$.
My Proof:
Let's say we have the example: $gbbgggbbgbb$.
We can visualise the solution using the above. With my example, I am going to add two bins to each side. This makes: $$\_ \, g\, b\, b\, g\, g\, g\, b\, b\, g\, b\, b\, \_$$ Let's first count boys to the left as we move along each girl.
Each string of boys or girls are either named $g_n$ or $b_n$.
- As we move along each girl, we get: $$g_1 \times b_1 + g_2(b_1 + b_2) + g_3(b_1 + b_2 + b_3) \ldots g_r(b_1 + b_2 + b_3 + \ldots b_r)$$
- As we move along each boy we get: $$b_1 \times g_1 + b_2(g_1 + g_2) + b_3(g_1 + g_2 + g_3) \ldots b_r(g_1 + g_2 + g_3 \ldots g_r)$$
For the girls we can write this sum out as: $$\begin{matrix} g_1 & & & & & & \color{teal}{b_1} \\ g_1 & g_2 & & & & & \color{teal}{b_2} \\ g_1 & g_2 & g_3 & & & & \color{teal}{b_3} \\ g_1 & g_2 & g_3 & g_4 & & & \color{teal}{b_4} \\ \vdots & & & & & & \color{teal}{\vdots} \\ g_1 & g_2 & g_3 & g_4 & \cdots & g_r & \color{teal}{b_r} \end{matrix}$$ Let's do the same for the boys: $$\begin{matrix} b_1 & & & & & & \color{teal}{g_1} \\ b_1 & b_2 & & & & & \color{teal}{g_2} \\ b_1 & b_2 & b_3 & & & & \color{teal}{g_3} \\ b_1 & b_2 & b_3 & b_4 & & & \color{teal}{g_4} \\ \vdots & & & & & & \color{teal}{\vdots} \\ b_1 & b_2 & b_3 & b_4 & \cdots & b_r & \color{teal}{g_r} \end{matrix}$$ Then take the diagonals of each diagram: \begin{align} &b_1 \times \color{teal}{g_1} + b_2 \times \color{teal}{g_2} + b_3 \times \color{teal}{g_3} + b_4 \times \color{teal}{g_4} + \ldots + b_r \times g_r \\ = \, &\color{teal}{g_1} \times b_1 + \color{teal}{g_2} \times b_2 + \color{teal}{g_3} \times b_3 + \color{teal}{g_4} \times b_4 + \ldots + \color{teal}{g_r} \times b_r \end{align} The above are both equal, so all other matching diagonals must be equal.
Therefore $G = B$
[Editor's note: this was copied verbatim from the original image here.]
I have answered this maths challenge problem.
The official solution is a lot less convoluted than mine, but I was wondering if I managed a full proof solution, or if there are any errors that I haven't noticed. Thanks for any help!