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Suppose we are to find the number of triangles that exist from the given figure enter image description here

I found one solution that says we let $A$ equal the number of internal lines from the top vertex, $B$ equal the number of internal lines parallel to the base, using the formula below to find the number of $N$ triangles $$N=\frac{(A+1)(A+2)(B+1)}{2}$$

With $A$ equal to 2 and $B$ equal to 3, we get $N=24$ triangles.

But can somebody explain why this formula works? How exactly do I derive this?

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2 Answers 2

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A triangle inside the figure has one of its sides being one of the $B+1$ horizontal lines, and the other two sides are two of the $A+2$ slant lines. So the total number of ways of forming a triangle is $\displaystyle {A+2 \choose 2} \times {B+1 \choose 1}$.

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  • $\begingroup$ Could you multiply the matrices together for me to show it getting 24? Because when I do it I keep getting 18 $\endgroup$
    – Lex_i
    Commented Jun 10, 2019 at 2:10
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    $\begingroup$ They are binomial coefficients, not matrices. $n \choose r$ is the number of ways of choosing $r$ objects from $n$ objects and the formula is ${n\choose r}=\frac{n!}{r!(n-r)!}$. In particular, ${A+2 \choose 2}=\frac{(A+2)(A+1)}{2}$ and ${B+1 \choose 1}=B+1$. $\endgroup$
    – CY Aries
    Commented Jun 10, 2019 at 2:14
  • $\begingroup$ ${n \choose 1}=\frac{n!}{(n-1)!1!}=\frac{n\times (n-1)!}{(n-1)!\times 1}=n$ and ${n \choose 2}=\frac{n!}{(n-2)!2!}=\frac{n(n-1)\times (n-2)!}{(n-2)!\times 2}=\frac{n(n-1)}{2}$. $\endgroup$
    – CY Aries
    Commented Jun 10, 2019 at 2:39
  • $\begingroup$ So ${A+2 \choose 2}=\frac{(A+2)(A+1)}{2}$ and $B+1 \choose 1$ is just $B+1$. $\endgroup$
    – CY Aries
    Commented Jun 10, 2019 at 2:51
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If we let $x$ to be the number of lines from the top vertex and $y$ to be the number of lines parallel to the base including the base, then we have $x = A+2$ and $y = B+1$. We keep this.

Now, think about how can we construct a triangle using these lines. First, we need a base, which we can choose from $y$ lines. Then, we need two other lines to construct a triangle but these two lines should intersect at one point (which is top vertex in this case). Therefore, we can actually choose $2$ lines from $x$ lines since all of these $x$ lines intersect at top vertex. So we can choose these $2$ lines with $\binom{x}{2} = \frac{x(x-1)}{2}$. So, in total, we can construct $\frac{x(x-1)y}{2}$ different triangles. Now, use $x = A+2$, $y = B+1$.

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