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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!

 Jul 13, 2022
 #1
avatar+118141 
+2

Please add a link to the previous question/answer

 Jul 13, 2022
 #3
avatar+12 
+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.

 Jul 14, 2022
 #4
avatar+118141 
+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

 Jul 15, 2022
edited by Melody  Jul 15, 2022
 #5
avatar+118141 
+1

A response from you, AngrySesame, would be good!

 Jul 16, 2022

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