Combinatorics: Counting quadrilaterals in triangle pattern

[Master1022]

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?

Attempt:
The video started with the simpler problem of finding the number of quadrilaterals in the following shape (a single 'strip' of the triangle):

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.

 

[pbuk]

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:

Spoiler: this should simplify to

$2n^2 - 3n + 1$

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.

 

[Master1022]

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:

Spoiler: this should simplify to

$2n^2 - 3n + 1$

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