+12 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!
+12 So, I got something for the first section!
There are three possible categories of arrangements for \((a, b, c)\). They are:
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.
+118608 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
