Let $n$ be a positive integer.

(a) Prove that

$n^3 = n + 3n(n - 1) + 6 \binom{n}{3}$

by counting the number of ordered triples $(a,b,c)$ of positive integers, where $1 \le a,b,c \le n,$ in two different ways.

What I did:

For this problem I just solved for the big equation which I thought would be pretty easy I just had to make the $n$ in the $\binom{n}{3}$ as $\dbinom{n \cdot (n-1) \cdot (n-2)}{3}$. Then I realized that the bottom part with $(a,b,c)$ was there, and I got stuck on that part.

What I'm stuck on:

I'm stuck on how to actually do it with the $(a,b,c)$ part. For example, should I prove the equation at the top and then deal with the part including $(a,b,c)$, and if so, what should I do with that part?

Thanks in advance I really need help on this :(

Guest Feb 23, 2021

#1**+1 **

Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face! And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c). If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks. There are n ways to choose a, and there are n - 1 ways to choose b. We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c. So there are 3n(n - 1) smirks.

Now we count the number of smiley faces. There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c. So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces. By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

Guest Feb 23, 2021