Let \(n\) be a positive integer.

(a) There are \(n^2\) ordered pairs \((a,b)\) of positive integers, where \(1≤a, b≤n.\) Using a counting argument, show that this number is also equal to \(n+2\binom{n}{2}.\)

(b) There are \(n^3\) ordered triples \((a,b,c)\) of positive integers, where \(1≤a, b,c≤n.\) Using a counting argument, show that this number is also equal to \(n + 3n(n - 1) + 6 \binom{n}{3}.\)

Please add an explanation. Thanks.

Guest Jun 21, 2023

#1**0 **

(a) Just expand both sides.

(b) 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).

Guest Jun 21, 2023