Combinatorics: Counting quadrilaterals in triangle pattern

AI Thread Summary
The discussion focuses on counting the number of quadrilaterals formed within a triangle pattern using combinatorial methods. An initial calculation yielded 36 quadrilaterals from a simpler shape, prompting the user to explore extending this concept to a larger triangle. The conversation emphasizes the need to consider different heights and orientations of quadrilaterals, suggesting a more complex approach than initially anticipated. Participants encourage further exploration of generalizing the findings for various rows and heights. The user expresses gratitude for the guidance and plans to refine their approach based on the feedback received.
Master1022
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Homework Statement
How many quadrilaterals are there?
Relevant Equations
Combinatorics
Hi,

I was watching a Youtube on combinatorics (here) and a problem was posed at the end of the video about counting the number of quadrilaterals.

Question:
How many quadrilaterals are present in the following pattern?
Screen Shot 2021-09-04 at 8.08.35 PM.png


Attempt:
The video started with the simpler problem of finding the number of quadrilaterals in the following shape (a single 'strip' of the triangle):
Screen Shot 2021-09-04 at 8.09.53 PM.png


The way I thought about it was that the quadrilateral was formed by taking two horizontal lines and two diagonal lines:
\begin{pmatrix} 2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 2 \end{pmatrix}

However, the individual triangles need to be deducted from the above product: there are ## = 2(5) - 1 = 9 ## individual triangles. Thus, I have:
\begin{pmatrix} 2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 2 \end{pmatrix} - 9 = 45 - 9 = 36 quadrilaterals

Now that I completed that basic problem, I am thinking how to extend that concept to the larger triangle case. I am trying to avoid a laborious method of looking at each row doing lots of sums because I hope that there is an elegant solution.

Currently, all I can think of is doing is:
\begin{pmatrix} \text{number of small triangles} \\ 2 \end{pmatrix} = \begin{pmatrix} 25 \\ 2 \end{pmatrix}.

This is because we can pick two triangles and form a quadrilateral between them.

Does this seem like it is headed along the correct path?

Any help is greatly appreciated.
 
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Master1022 said:
Does this seem like it is headed along the correct path?
It started well but I think it is in danger of heading off course.

You have successfully calculated the number of quadrilaterals of height 1 which have their base in the 5th row, and with a bit more work you can generalise this to the number of height 1 which have their base in the n'th row:
## 2n^2 - 3n + 1 ##
You now need to look at how many quadrilaterals of height 2 have their base [edit: or bottom vertex] in the 5th row. See if you can generalise this expression to how many quadrilaterals of height h have their base [edit: or bottom vertex] in the n'th row. [edit: This will not be simple - there are many different orientations to consider.] You then need to sum over appropriate ranges of h and n to get to the total.
 
Last edited:
pbuk said:
It started well but I think it is in danger of heading off course.

You have successfully calculated the number of quadrilaterals of height 1 which have their base in the 5th row, and with a bit more work you can generalise this to the number of height 1 which have their base in the n'th row:
## 2n^2 - 3n + 1 ##
You now need to look at how many quadrilaterals of height 2 have their base [edit: or bottom vertex] in the 5th row. See if you can generalise this expression to how many quadrilaterals of height h have their base [edit: or bottom vertex] in the n'th row. [edit: This will not be simple - there are many different orientations to consider.] You then need to sum over appropriate ranges of h and n to get to the total.
Thanks @pbuk for the reply! Yes, I think I tried to over-simplify the problems. Thanks for the hints, I will give them some thought and post my next attempts here soon. Many thanks
 
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