SELECTION EXAMINATION

We suppose that the square is placed into a Cartesian system of coordinates with origin $A(0,0)$ and axes on the sides $AB$ and $AD$, as in the figure 7. Let the spider start from $A$ and make its first movement ($m$ steps right and $k$ steps up or $m$ steps up and $k$ right). In the figure you can see the case $m=4,k=3$.



After the first movement the spider can be on the point $M(k,m)$ or on the point $N(m,k)$. The number of ways the spider can approach the point $M(k,m)$ or the point $N(m,k)$ are equal to the combinations of $k+m$ elements taken $k$ at a time, that is $$\left(\genfrac{}{}{0ex}{}{k+m}{k}\right)=\left(\genfrac{}{}{0ex}{}{k+m}{m}\right)=v.$$

After the second movement the spider can be at three different points: $K(2k,2m)$, $L(k+m,k+m)$, and $T(2m,2k)$ with corresponding number of ways $v^2$, $2v^2$, and $v^2$. The point $M$ can be approached by $v$ different ways from $A$. The point $K$ can be approached by $v$ different ways from $M$. Hence the point $K$ can be approached by $v^2$ different ways from point $A$.



In a similar fashion we can find that point $T$ can be approached by $v^2$ different ways from point $A$. The point $L$ can be approached by $v^2$ different ways by following the route $A,M,L$ and by $v^2$ different ways by following the route $A,N,L$. Hence point $L$ can be approached by $2v^2$ different ways from point $A$.

Similarly, we conclude that after the third movement the spider can be placed in four different points that can be approached by $\left(\genfrac{}{}{0ex}{}{3}{3}\right)\cdot v^3$, $\left(\genfrac{}{}{0ex}{}{3}{2}\right)\cdot v^3$, $\left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot v^3$, and $\left(\genfrac{}{}{0ex}{}{3}{0}\right)\cdot v^3$ different ways, respectively. Therefore, after the completion of the $l$ movements, the spider will be placed at $l+1$ different points, lying on the line with equation: $x+y=l(k+m)$.



If $r=m-k$, then these points are the following: $$A_0(lm,lk),A_1(lm-r,lk+r),A_2(lm-2r,lk+2r),\dots ,A_l(lm-lr,lk+lr)\equiv A_l(lk,lm).$$ These points can be approached by $\left(\genfrac{}{}{0ex}{}{l}{l}\right)\cdot v^l$, $\left(\genfrac{}{}{0ex}{}{l}{l-1}\right)\cdot v^l$, $\left(\genfrac{}{}{0ex}{}{l}{l-2}\right)\cdot v^l$, $\dots$, and $\left(\genfrac{}{}{0ex}{}{l}{0}\right)\cdot v^l$ different ways from the point $A$. The point $C(ml,ml)$ can be approached from the point:

$$A_0(lm,lk),\text{ with }\left(\genfrac{}{}{0ex}{}{(ml-lm)+(ml-lk)}{ml-lk}\right)=\left(\genfrac{}{}{0ex}{}{ml-lk}{ml-lk}\right)=\left(\genfrac{}{}{0ex}{}{lr}{lr}\right)=1,\text{ way.}$$ $$A_1(lm-r,lk+r),\text{ with }\left(\genfrac{}{}{0ex}{}{(ml-lm+r)+(ml-lk-r)}{r}\right)=\left(\genfrac{}{}{0ex}{}{ml-lk}{r}\right),\text{ ways.}$$ $$A_2(lm-2r,lk+2r),\text{ with }\left(\genfrac{}{}{0ex}{}{(ml-lm+2r)+(ml-lk-2r)}{2r}\right)=\left(\genfrac{}{}{0ex}{}{ml-lk}{2r}\right),\text{ ways.}$$ $$A_l(lk,lm),\text{ with }\left(\genfrac{}{}{0ex}{}{(ml-lk)+(ml-lm)}{ml-lk}\right)=\left(\genfrac{}{}{0ex}{}{lr}{lr}\right)=1,\text{ way.}$$ Hence the point $C(ml,lm)$ can be approached from point $A$ by: $$v^l\left(\left(\genfrac{}{}{0ex}{}{l}{l}\right)\cdot \left(\genfrac{}{}{0ex}{}{lr}{0}\right)+\left(\genfrac{}{}{0ex}{}{l}{l-1}\right)\cdot \left(\genfrac{}{}{0ex}{}{lr}{r}\right)+\left(\genfrac{}{}{0ex}{}{l}{l-2}\right)\cdot \left(\genfrac{}{}{0ex}{}{lr}{2r}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{l}{0}\right)\cdot \left(\genfrac{}{}{0ex}{}{lr}{lr}\right)\right)\text{ ways.}$$

(*) Moving from the point $(i,j)$ to the point $(i+k,j+m)$ of the grid with direction right and up, can be made by $\left(\genfrac{}{}{0ex}{}{k+m}{k}\right)=\left(\genfrac{}{}{0ex}{}{k+m}{m}\right)$ different ways. The procedure of finding the number of ways is the same as the procedure we have used at the first movement.

Note. In the case where $m-k=r=1$, then the number of all possible ways becomes: $$v^l\left(\left(\genfrac{}{}{0ex}{}{l}{l}\right)\cdot \left(\genfrac{}{}{0ex}{}{l}{0}\right)+\left(\genfrac{}{}{0ex}{}{l}{l-1}\right)\cdot \left(\genfrac{}{}{0ex}{}{l}{1}\right)+\left(\genfrac{}{}{0ex}{}{l}{l-2}\right)\cdot \left(\genfrac{}{}{0ex}{}{l}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{l}{0}\right)\cdot \left(\genfrac{}{}{0ex}{}{l}{l}\right)\right)=v^l\left(\genfrac{}{}{0ex}{}{2l}{l}\right)={\left(\genfrac{}{}{0ex}{}{k+m}{m}\right)}^l\left(\genfrac{}{}{0ex}{}{2l}{l}\right).$$