In how many ways can the elements of [n] be permuted so that the sum of every two consecutive elements in the permutation is odd?
[n] is the set of integers from 1 to n.
My attempt:
If n is even:
If n is odd, we must start with an odd element:
\left(\frac{n-1}{2}\right)\left(\frac{n-1}{2}\right)\left(\frac{n-3}{2}\right)\left(\frac{n-3}{2}\right)...)
Is this right?
I agree with you as far as that goes.Originally Posted by alexmahone
In how many ways can the elements of [n] be permuted so that the sum of every two consecutive elements in the permutation is odd?
[n] is the set of integers from 1 to n.
My attempt:
If n is even:
If n is odd, we must start with an odd element:
But in is odd it should be: because start with an odd and alternate
![\left[ {\left( {\frac{{n + 1}}{2}} \right)!} \right] \cdot \left[ {\left( {\frac{{n - 1}}{2}} \right)!} \right]](http://latex.codecogs.com/png.latex?\left[ {\left( {\frac{{n + 1}}{2}} \right)!} \right] \cdot \left[ {\left( {\frac{{n - 1}}{2}} \right)!} \right])
Actually, that's exactly what I got. But there was a problem with the latex (which I've now fixed).I agree with you as far as that goes.
But in is odd it should be: because start with an odd and alternate
\left(\frac{n-1}{2}!\right)^2 = \left(\frac{n+1}{2}!\right)\left(\frac{n-1}{2}!\right))
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