## Permutations with given sign

How many permutations on $$n$$ elements have sign 1, and how many sign $$-1$$?

• #### Resolution

Altogethere there are $$n!$$ permutations on $$n$$ elements.

Permutations with the positive sign form a subgroup. For $$n\ge 2$$ exist also permutations with the negative sign (odd). These form the factorization class different from the subgroup.

An alternative elementary argument:

From a permutation $$p$$ on $$n-1$$ elements can be formed a permutation on $$n$$ elements by putting $$n$$ on the first, second, etc., the last $$n$$-th position. In the first case will newly appear $$n-1$$ inversion, in the second $$n-2$$, etc. in the last case there is no new inversion. hence $$\lceil\frac{n}{2}\rceil$$ of the new permutations has the same sign as $$p$$ and $$\lfloor\frac{n}{2}\rfloor$$ has the opposite sign. This solves the case for an even $$n$$ – there are as many odd as even permutations.

For odd $$n\ge 3$$ we must use the assumption on the same number of odd and even permutations on $$n-1$$ elements, that is $$\frac{(n-1)!}{2}$$.

Then there are $$\frac{(n-1)!}{2}\lceil\frac{n}{2}\rceil+ \frac{(n-1)!}{2}\lfloor\frac{n}{2}\rfloor= \frac{n!}{2}$$ even (and also odd) permutations.

• #### Result

Permutations with the positive sign are as many as those with the negative sign, i.e. $$\frac{n!}{2}$$ (for $$n\ge 2$$).