GRE Question [Word Problem]

Question

The following is a question i got wrong on the GRE practice test. There is no explanation provided. I have actually a confusion as to what is even being asked and how they get the answer $r^{2} - r$. Can anyone please provide what approach to take ?.

Answer 1

The array is $r \times (r+1)$, so has $r^2+r$ cells. Striking out a row eliminates $r+1$ of these. Striking out a column eliminates $r$ of these. The cell at the intersection of the row and column has been struck twice, but should only be deleted once, so we add one back in. $r^2+r-(r+1)-r+1=r^2-r$

Answer 2

The total number of squares on the board is $r(r+1)$. There are $r$ squares in the seventh column, $(r+1)$ squares in the fourth row, and $1$ in both. So the answer is

$$T = r(r+1) - (r + r+1 - 1) = r^2 + r - 2r = r^2 - r.$$

Answer 3

$\newcommand{\ss}{\color{white}\cdot}$Since there are $r+1$ columns, each row has $r+1$ squares; in particular, the $4$-th row has $r+1$ squares. Since there are $r$ rows, each column has $r$ squares; in particular, the $7$-th column has $r$ squares. Thus, there are $r+1$ squares in the $4$-th row and $r$ squares in the $7$-th column. On the face of it that seems to say that there are $(r+1)+r=2r+1$ squares that are in the $4$-th row or the $7$-th column. However, one square is in both the $4$-th row and the $7$-th column, so we’ve counted it twice, and there are really only $2r$ distinct squares that are in the $4$-th row, the $7$-th column, or both.

The board as a whole has $r(r+1)=r^2+r$ squares, and we’ve just seen that $2r$ of those squares are in the $4$-th row or $7$-th column (or both); that leaves

$$(r^2+r)-2r=r^2-r$$

squares that are in neither the $4$-th row nor the $7$-th column.

A picture may help: here’s what the board looks like when $r=4$.

$$\begin{array}{|c|c|c|c|c|} \hline \ss&\ss&\ss&\bullet&\ss\\ \hline \bullet&\bullet&\bullet&\color{red}\bullet&\bullet\\ \hline \ss&\ss&\ss&\bullet&\ss\\ \hline \ss&\ss&\ss&\bullet&\ss\\ \hline \end{array}$$

I’ve used bullets to mark the squares in the second row and in the fourth column. You can see that there are $r+1=5$ squares in the second row and $r=4$ in the fourth column, for a total of $5+4=9$, but the red bullet is counted twice, so there are actually only $8=2r$ squares that are either in the second row or the fourth column (or both).