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Let n be a positive integer.
(a) There are n2 ordered pairs of positive integers, where 1≤a,b≤n. Using a counting argument, show that this number is also equal to
n+2(n2)
(b) There are n3 ordered triples 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(n3).
Here is the answer to part (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).
