Explanation

If we were just counting all of the squares in the board game we could use multiplication: r rows times r+1 columns.

This would give us \(r^2+r\) squares.

If we take one row out (doesn't matter if its the first, second, third, etc) then we would have r-1 rows times r+1 columns.

If we also then take a column out, we would have r-1 rows times r columns.

So the total number of squares without a column and without a row is \(r*(r-1)=r^2-r\).

So option A is correct.
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what if we calculate all the cells in total first likeso r*(r +1) = r^2 + r

and then calculate the total cells missing likeso r + r + 1 = 2r + 1

and then substract one value from another

likeso r^2 + r - 2r - 1 = r^2 -r -1 here we have one extra one. Why is that?

nevermind... got it

NOTE: After creating the graphics, I see that I used n instead of r.
Please forgive me!


Let's start with an n by n+1 grid

The number of squares = (n)(n + 1) = n² + n

In the 4th row, we can see there are n+1 squares


In the 7th column, there are n squares.

HOWEVER one of the squares is already shaded blue.
So, there are n-1 red squares

TOTAL number of UNSHADED squares = n² + n - (n + 1) - (n - 1)
= n² - n

Answer: A

Cheers,
Brent
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Can you please explain how you resolved the extra 1? I'm running into the same problem

The main issue is highlighted above.

If the total number of missing cells = r + (r + 1), then we are counting the shared cell TWICE.
So, we must subtract 1 from this sum.
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Can also solve using values. Is this approach wrong?

We are given that r>10 so we can say rows=11 and columns=12. Therefore total boxes= 12*11=132 and boxes in 4th row AND 7th column= 12+11–1(common to both row and column)=22.

Now what's important to remember is that we are asked other rows, which as per our numbers is 132–22=110. Now we need to input r=11 for the options and see which options output 110. A does so => 11^2–11= 121–11=110.

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