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$$\frac {n-2}{n} \cdot \frac {n-3}{n-1} \cdot \frac {n-4}{n-2} \cdots \frac{2}{4} \cdot \frac{1}{3} = \frac {1}{n(n-1)}$$

Why is this true? Notice the denominators and numerators cancel out, but since they "aren't in sync" the first two denominators and the last two numerators will not be cancelled out. Considering this, shouldn't the product be:

$$\frac{2}{n(n-1)}$$

What am I misunderstanding?

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1 Answer 1

up vote 4 down vote accepted

Yes, it should be $$\dfrac2{n(n-1)}$$ To see this take $n = 4$, we then get that $$\dfrac24 \cdot \dfrac13 = \dfrac{2}{4(4-1)}$$

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So it's not just a misunderstanding, the book is wrong, correct? –  user26649 Jul 9 '12 at 4:29
    
@Riddler Yes, unless you have copied something wrong. –  user17762 Jul 9 '12 at 4:29
    
I double checked it -I guess it's an error . Thanks for the help Marvis! –  user26649 Jul 9 '12 at 4:44

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