2
$\begingroup$

Set $n \ge 1$. Prove that $$n!-1 = \sum_{X \in \{1,2,...,n-1\}^P} \prod_{y \in X} y.$$

I found this one while studying my combinatorics lecture notes. Intuitively, $n!-1=(1+1)\cdot(2+1)\cdots(n-1+1)-1$. Expand and it looks like the resulting sum of products. I tried to prove this identity using induction but it somehow got very complicated.

Remark: $P=$ powerset.

  • 1
    You're on the right track. You merely have to prove "Expand and it looks like the resulting sum of products". However, this may be simpler than you imagine...(Think of Vieta's Formula) – 2017-02-25
  • 0
    Why the $-1$? The sum equals $n!$. It seems the setters are ignoring the empty set. – 2017-02-25
  • 0
    It is generally noted $X\in\mathcal P(\{1,2,..,n-1\})$ for powerset, so that we do not confuse and get result $P\times(n-1)!$ – 2017-02-25
  • 0
    @zwim In fact it has many notations, for example $2^{\{1,\ldots,n-1\}}$. [Wikipedia](https://en.wikipedia.org/wiki/Power_set) lists even more, though they are variants of your notation. – 2017-02-25
  • 1
    @Yuval Maybe but for me it would be easier to do math using my french words, but I use english terms so that most of SE people understand, I think it goes along with the notations, better use the more widely accepted ones. – 2017-02-25

2 Answers 2

1

You can prove this by induction (with $n!-1$ replaced by $n!$). The idea is that $$ \sum_{X \in \{1,\ldots,n\}^P} f(X) = \sum_{X \in \{1,\ldots,n-1\}^P} [f(X) + f(X \cup \{n\})]. $$ In your case $f(X)$ is the product of all elements in $X$ (if any).

0

Think about the polynomial $\prod_{i=1}^n(x+i)$. In expanding each factor in the product, you have to choose either an $x$ or a number. So, $$ \prod_{i=1}^n(x+i)=\sum_{X\subseteq \{1,\ldots,n\}}x^{n-|A|}\prod_{y\in X}y.$$ Your result follows by letting $x=1$.