+0

# I'm quite stuck here. Anyone?

0
467
5
+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 $1, 2, 3, \dots, n + 2$ containing three different numbers in two different ways.

Thank you!

Jul 13, 2022

#1
+118577
+2

Jul 13, 2022
#2
+12
+1
Jul 14, 2022
#3
+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
+118577
+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
+118577
+1

A response from you, AngrySesame, would be good!

Jul 16, 2022