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As shown in the figure above 20 equal squares are arranged to form a
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Updated on: 03 Dec 2015, 21:34
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Jamboree way
If we choose any two horizontal lines and any two vertical lines the enclosed quadrilateral formed will be a rectangle. The given figure has 5 horizontal lines and 6 vertical lines. The number of ways we can select 2 horizontal lines out of 5 horizontal lines = 5C2 = 10 The number of ways we can select 2 vertical lines out of 6 vertical lines = 6C2 = 15
Total number of rectangles formed = (The number of ways we can select 2 horizontal lines) x (The number of ways we can select 2 vertical lines)
Total number of rectangles formed = 10 x 15 = 150
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12 Oct 2015, 00:23
Hi Aryama,
could you kindly confirm if this type of question is likely to appear on the GMAT? I had not seen before anything like it in any of the CATs I've taken, or around here.
And is there a "formula"-way to solve it? Using combinatorics, maybe? Or are we supposed to simply count?
Many thanks in advance!
Paula
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12 Oct 2015, 11:59
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I don't think that such kind of problems appear in GMAT. We can use this formula to solve this one: \(\frac{m(m+1)*n(n+1)}{4}\)==> \(\frac{4*5*5*6}{4}\)=150
Alternative solution: In a rectangle m × n there are (m+1) vertical grid lines and (n+1) horizontal grid lines (5 and in the example here).
To define any rectangle within the grid, we must choose 2 of each and there are ( (m+1) choose 2 ) × ( (n+1) choose 2 ) ways to do that.
For 5 × 6 that gives us 2C5 × 2C6 = 10 * 21 = 150
What would the experts say about this question, isn't it too specific/hard for GMAT ?
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As shown in the figure above 20 equal squares are arranged to form a
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03 Dec 2015, 04:28
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AryamaDuttaSaikia wrote:
Jamboree way
If we draw two horizontal lines and two vertical lines the enclosed quadrilateral formed will be a rectangle.
The given figure has 4 horizontal lines and 3 vertical lines. The number of ways we can select 2 horizontal lines out of 4 horizontal lines = 4C2 = 6
The number of ways we can select 2 vertical lines out of 3 vertical lines = 3C2 = 3 Total number of rectangles formed = (The number of ways we can select 2 horizontal lines) x (The number of ways we can select 2 vertical lines)
Total number of rectangles formed = 6 x 3 = 18
Hi, I think you require to change your answer as jamboree way, which is also the normal way, is correct but seems to be answering some other Q....
there is a grid of 6 by 5 and not 4 by 3.. for a rectangle we can choose 2 horizontal points and 2 vertical points... total ways = 6C2 * 5C2=15*10=150
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For a rectangle we need two horizontal lines out of 5 and we need 2 vertical lines out of 6
= 6C2*5C2 = 15*10 = 150
Answer: option D
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18 Jun 2017, 11:50
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but here the question asks for rectangles . I think this no 150 includes even squares also . like vertical grid line 2and 4 will give a square in combination with horizontal lines( 1,3) ; (2,4) ; (3,5) and so on should we not exclude them from this no .
As shown in the figure above 20 equal squares are arranged to form a
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04 Jan 2018, 15:23
meenakshibehera wrote:
but here the question asks for rectangles . I think this no 150 includes even squares also . like vertical grid line 2and 4 will give a square in combination with horizontal lines( 1,3) ; (2,4) ; (3,5) and so on should we not exclude them from this no .
You are right. The solution presented includes square as well. The formulae is blindly applied to include squares. To prove, we can solve a simple combination. Let's take a 2X3 rectangle grid made up of small squares. Applying the same logic, it will be 3C2 X 4C2 = 3*6 = 18. This will include squares as well.
There are only 10 such rectangles which can be formed here. rectangles with 1x2 = 2 per row = 4 in total rectangles with 2x1 = 1 per column = 3 in total rectangles with 1x3 = 1 per row = 2 in total rectangle with 2x3 = 1 in total
Moreover, there are 8 such squares in total squares 1x1 = 2*3 = 6 squares 2x2 = 1*2 = 2
Now, applying the same to the question. Total quadrilaterals that can be formed is 5C2 * 6C2 = 10*15 = 150. The number of squares in this will include = 4*5 + 3*4 + 2*3 + 1*2 = 40 Hence, the total number of rectangles will be 150 - 40 = 110
With respect to the question, I suppose (looking at the picture), the smaller units appear to be rectangle. Hence, the question was incorrectly worded as 20 equal squares instead of 20 equal rectangles. If it is the latter, 150 is right.
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20 Oct 2018, 11:57
With respect to the question, I suppose (looking at the picture), the smaller units appear to be rectangle. Hence, the question was incorrectly worded as 20 equal squares instead of 20 equal rectangles. If it is the latter, 150 is right.[/quote]
I don't think that there's any problem with the wording of the question since 'A square is also a rectangle'.
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27 Aug 2019, 06:56
AryamaDuttaSaikia wrote:
Jamboree way
If we choose any two horizontal lines and any two vertical lines the enclosed quadrilateral formed will be a rectangle. The given figure has 5 horizontal lines and 6 vertical lines. The number of ways we can select 2 horizontal lines out of 5 horizontal lines = 5C2 = 10 The number of ways we can select 2 vertical lines out of 6 vertical lines = 6C2 = 15
Total number of rectangles formed = (The number of ways we can select 2 horizontal lines) x (The number of ways we can select 2 vertical lines)
Total number of rectangles formed = 10 x 15 = 150
This approach will also include squares formed by selecting 2 horizontal and 2 vertical lines(if distance between horizontal lines is equal to distance between vertical lines) so answer should definitely be less than 150.
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28 Aug 2019, 07:33
meenakshibehera wrote:
but here the question asks for rectangles . I think this no 150 includes even squares also . like vertical grid line 2and 4 will give a square in combination with horizontal lines( 1,3) ; (2,4) ; (3,5) and so on should we not exclude them from this no .
Every square is a rectangle, but every rectangle is NOT a square. A rectangle having its length equal to width is a square.
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47% (02:01) correct 
