Question

Prove that the product of 2n consecutive negative integers is divisible by (2n)!

Answer 1

Let  

\[2n\] negative integers be 

\[\left( - r \right), \left( - r - 1 \right), \left( - r - 2 \right), . . . . , . . . , \left( - r - 2n + 1 \right)\]

Then, product = \[\left( - 1 \right)^{2n} \left( r \right)\left( r + 1 \right)\left( r + 2 \right), . . . . , . . . \left( r + 2n - 1 \right)\]

\[= \frac{\left( r - 1 \right)! \left( r \right)\left( r + 1 \right) \left( r + 2 \right) . . . . . . \left( r + 2n - 1 \right)}{\left( r - 1 \right)!}\]
\[ = \frac{\left( r + 2n - 1 \right)!}{\left( r - 1 \right)!}\]
\[ = \frac{\left( r + 2n - 1 \right)!}{\left( r - 1 \right)!\left( 2n \right)!} \times \left( 2n \right)!\]
\[ = {}^{r + 2n - 1} C_{2n} \times \left( 2n \right)!\]

This is divisible by 

\[\left( 2n \right)! .\]