Probability of Rectangle - Big Confusion!!!

[gmatNooB8787]

Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

(Note that two rectangles that have the same four vertices that are labeled differently are considered to be the same rectangle.)

Choices
A 396
B 1,260
C 1,980 <- Explained Answer
D 7,920 <- My Answer
E 15,840

Given Soln : It says we can choose 2 x coordinates and 2 y coordinates
So answer is : 9C2 * 11C2 = 1980

My Soln: However I solved this according to the method given in OG12.
A---------B
| |
| |
C---------D

So the possibilities for C(x,y) = 9*11

For D : D(y) same as C(y) and D(x) can be any of the other options
hence D(x,y) = 8 * 1;

Similarly for B(x,y) = 1 * 10
For A(x,y) = 1*1

Hence total Rectangles will be : 9*11*8*10 = 7920.

This is the exact method used in the OG. So , Which Solution is correct ? Please Help.

[Mike@Magoosh]

Dear gmatNooB8787,

I'm happy to help clarify this.

First of all, the answer of (C) 1980 is the correct answer. The way they explain (9C2 * 11C2 = 36*55 = 1980) is how I would solve it --- if you don't understand that approach, I will happily explain it, because combinations (nCr) are an excellent topic to know for the GMAT. Here's a blog I wrote you can read about that:
https://magoosh.com/gmat/2012/gmat-permu ... binations/
Let me know if you would like even more explanation.

Now, you followed what you understood as the OG's approach, and you got a different answer. Why?Well, let's go through that carefully.

The possibilities for C(x,y) = 9*11 = 99. Quite true.

D can be anything else on that horizontal line, so there are 8 possibilities for D, and thus, 8*9*11 possibilities for CD. BUT, we have to pause here and be careful --- for example, for the pair (4, 1) and (7, 1), a valid pair for CD, we could have come up with that same pair two different ways --- it could have arisen with C = (4, 1) and D = (7, 1), or as C = (7, 1) and D = (4, 1). Thus, that number, 8*9*11, counts every pair twice, so we need to divide it by 2 --- that would be 4*9*11.

Now, pick a y-coordinate for A --- there are 10 possibilities --- and that completely determines B. BUT AGAIN, if we pick CD on y = 1, and AB on the line y = -3, that produces the same rectangle as picking CD on y = -3 and AB on y = 1. Again, we have counted every rectangle twice, so we have to divide by 2 again --- instead of 10, we divide down to 5, and then multiply.

Lo and behold ---> 5*4*9*11 = 1980

Your answer was four times too big, because in two different ways you counted points twice. That's what's hard about applying the Fundamental Counting Principle correctly ---- you always have to have your eye out for how you might be creating duplicates in your counting. That's why combinations are much much easier, because that approach automatically omits duplicates, so you don't have to worry about them.

I hope this was helpful. Please let me know if you have any more questions.

Mike