Let 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.

For part A, I understand we can create cases for how a, b, and c are relative to each other. I don't know what else to do though :(

(b) 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.

Guest Aug 2, 2020

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Guest
Aug 2, 2020

#1**0 **

(a) 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).

(b) To choose three numbers, we can choose two groups one group with two numbers and the other group has one number. The total of n + 2 numbers can be separated into two groups.

In the first way, one group has 2 numbers, and the other group has n numbers. They form a total of n + 2 numbers. There are C(2,2) = 1 ways to choose two numbers from the 2 group. There are C(n,1) = n ways to choose one number from the one group. This gives us a first term of 1*n.

In the second way, one group has three numbers, and the other group has n - 1 numbers. They form a total of n - 1 numbers. There are C(3,2) = 2 ways to choose two numbers from the 2 group. There are C(n - 1,1) = n - 1 ways to choose one number from the one group. This gives us a second term of 2*(n - 1). The pattern will continue until we reach n. So the two sides are equal.

Guest Aug 14, 2020