Let \(n\) 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.
(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.
https://web2.0calc.com/questions/need-some-help-please_36
In that answer by another guest they said that when you plot the equation you get a smiley face and some other faces when you change it up. I was confused about how that worked, could someone please explain it to me? The question was locked so I couldn't ask on there. Thank you!
+874 https://web2.0calc.com/questions/please-melody-or-alan-help-i-am-stuck-and-please#r11
Induction works perfectly.
+874 Ideally, what (s)he meant by smiley faces were when $b < a \leq c | b < c \leq a$, and frowny faces were if $b > a \geq c | b > c \geq a$.
Neutral face is when $a = b = c$. Smirks are all other possibilities of $a, b, c.$
Number of neutral faces: $n$ ways to choose $a$, and since $a=b=c,$ there are $n$ ways.
Number of smirks: $\binom{3}{2} \cdot n(n-1)$ because there are $n$ ways to choose $a,$ $n-1$ ways to choose $b$ or $c,$ and you could have any one of these be first so you multiply $n(n-1)$ by $\binom{3}{2}$ or $\binom{3}{1},$ either is same. So you have $3(n)(n-1)$ ways.
Number of smiley faces: $P\binom{n}{3} = 3 \cdot \binom{n}{3}.$
Number of frowny faces: Symmetric to number of smiley faces = $3 \cdot \binom{n}{3}.$
Number of ways to choose $a, b, c:$ $n \cdot n \cdot n = n^3.$
Add these up and you get $n^3 = n + 3n(n - 1) + 6 \binom{n}{3}$
