1. Give a combinatorial proof

n! 2(n 1)! = (n 1)(n 2)(n 2)!

I don't understand how to even start, an algebraic proof isn't an acceptable answer

2. Take into account that is $n!= n\ (n-1)!$, so that $(n-1)!$ is 'common factor'...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by KrazEcat01
n! 2(n 1)! = (n 1)(n 2)(n 2)!

I don't understand how to even start, an algebraic proof isn't an acceptable answer
A combinatorical proof will also involve algebra. $\displaystyle \chi \sigma$'s answer is the best...

4. The LHS is the number of permutations of $\{1,...,n\}$ such that 1 ends up neither in the first nor in the second position.
( $n!$ permutations in all, $(n-1)!$ permuations with 1 in the first position, and $(n-1)!$ permutations with 1 in the second position , hence $n!-2(n-1)!$ permutations ...)

So let's count this in a different way, first, let's choose a number for the first position, we have $(n-1)$ choices since we can't choose 1, for the second position we have $(n-2)$ choices -since we can't choose 1 and we can't the number that we have chosen for the first position.
To complete the rest we have $(n-2)!$ ways, thus: $(n-1)\cdot (n-2)\cdot (n-2)!$ ways in all.