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How could we show that: $\prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{(n+1)!}{(n+1-i)\cdot i! \cdot (n-i)!}(-1)^{(n-i)} .$

The module suggest we could reduce it by simply writing \prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{(n + 1)n \cdots (n + 1 − (i − 1))(n + 1 − (i + 1)) \cdots 1}{i(i − 1) \cdots 1 · (−1) \cdots (−(n − i))} , but I am not able to figure out the result from here.

Please explain your answer.

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    @Yuqing:You are right,my bad typo.2011-05-30

3 Answers 3

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The numerator is

$\prod_{0 \le j \ne i \le n} (n+1-j) \,.$

By writing it term by term you get

$(n+1)n(n-1)....(n+1-(i-1))(n+1-(i+1))...1 = \frac{(n+1)!}{n+1-i} \,.$

The denominator is

$\prod_{0 \le j \ne i \le n} (i-j)= [i(i-1)(i-2)...(i-(i-1))][(i-(i+1))(i-(i+1))...(i-n)] \,.$

The first bracket is exactly $i!$, while the second is $(-1)(-2)...(-(n-i))$. By taking a minus out of each bracket you get $(-1)^{n-i}(n-i)!$.

Thus the denominator is $(-1)^{n-i}(n-i)!i!$.

Combining those you get exactly the formula Yuqing posted.

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    Lol...clever ;)2011-05-30
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\begin{equation} \prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{\prod_{0 \le j \ne i \le n}(n+1-j)}{\prod_{0 \le j \ne i \le n}(i-j)} \end{equation}

we can rewrite this as follows

\begin{equation} \frac{\prod_{0 \le j \ne i \le n}(n+1-j)}{\prod_{0 \le j \ne i \le n}(i-j)} = \frac{\prod_{0 \le j \le n}(n+1-j)}{(n+1-i) \times \prod_{0 \le j < i}(i-j) \times \prod_{i+1 \le j \le n}-(j-i)} \end{equation}

You can simplify the above expression to get an expression similar to what you have written. However, there is only $(n+1-i)$ instead of of the $(n+1-i)!$ in the denominator on the right hand side.

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    @debanjan, I'm simply splitting the product into two terms, one where j, where the individual terms of the product \prod_{0\le j are all positive and another where the product is of negative terms. Try it with a small example (say $n=3$ and $i=1$) and it should be clear.2011-05-30
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I assume only $j$ is the running index. Typically the way you have written the product on the left side it means that $i$ and $j$ are both running indices. You could mention this explicitly by using the following notation. $\prod_{\substack{j=0\\ j \neq i}}^{n} \frac{n+1-j}{i-j}$ Split the product into two $\prod_{\substack{j=0\\ j \neq i}}^{n} \frac{n+1-j}{i-j} = \prod_{j=0}^{i-1} \frac{n+1-j}{i-j} \times \prod_{j=i+1}^{n} \frac{n+1-j}{i-j}$ $\prod_{j=0}^{i-1} \frac{n+1-j}{i-j} = \frac{(n+1)n(n-1)(n-2) \cdots (n-i+2)}{(i)!} = \frac{(n+1)!}{i! (n-i+1)!}$ $\prod_{j=i+1}^{n} \frac{n+1-j}{i-j} = (-1)^{n-i} \frac{(n-i)!}{(n-i)!} = (-1)^{n-i}$ Hence, we get $(-1)^{n-i} \frac{(n+1)!}{i! (n-i+1)!}$

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    Actually both i and j are running index in the actual problem,but you are right this is the better way to explain for this subproblem.2011-05-30