**NOTE: I know this has been solved before, but I didn't understand the previous user's solution, so I'm asking again. Thanks!**

Let n be a positive integer.

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 \(1 \leq a, b, c \leq n\)

Secondly,

Prove that

\(\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1),\)

by counting the number of subsets of [math]1, 2, 3, \dots, n + 2[/math] containing three different numbers in two different ways.

Thank you!

AngrySesame Jul 13, 2022

#3**+1 **

So, I got something for the first section!

There are three possible categories of arrangements for \((a, b, c)\). They are:

- \((a, b, c)\) are equal.
- \((a, b, c)\) are all different.
- Two variables are equal, one is not.

There are \(6\binom{n}{3}\) arrangements when all the variables are different. \(\binom{n}{3}\) accounts for all the arrangements regardless of order, but we must multiply by 6 to get our final answer.

There are \(n\) arrangements when all the variables are the same.

There are \(3n(n - 1)\) ways to compute the arrangements when two variables are the same, and one is not equal.

Now, I need to figure out b; which I cannot.

AngrySesame Jul 14, 2022

#4**+1 **

I have a solution.

I think it is what the other answerer was trying to explain. (Although I am not sure)

\(\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1)\)

Method 1:

The number of subsets of 3 elements from the set of n+2 elements is given by \(\binom{n+2}{3}\)

Method 2:

Let us have 2 sets, A(n elements) and B(2 elements)

How many ways are there to choose 3 elements if two MUST come from set B answer: **n**

Now take one of the elements from A and put it in set B so A(n-1 elements) and B(3 elements)

How many ways are there to choose 3 elements if two MUST come from set B but one of the 2 must be the new one answer: (n-1) 2 = **2(n-1)**

Now take one of the elements from A and put it in set B so A(n-2 elements) and B(4 elements)

How many ways are there to choose 3 elements if two MUST come from set B but one of the 2 must be the new one answer: (n-2) 3 = **3(n-2)**

....

and so on until all but one of the elements are in set B n(n-(n-1) =** n(1)**

So The number of subsets of 3 elements from the set of n+2 elements is also given by

\( (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1)\)

HENCE

\(\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1),\)

I have shown that they are equal by counting in 2 different ways

Melody Jul 15, 2022