Random Permutation Statistics - A Problem From The 2005 Putnam Competition

A Problem From The 2005 Putnam Competition

A link to the Putnam competition website appears in the section External links. The problem asks for a proof that

$\sum_{\pi\in S_n} \frac{\sigma(\pi)}{\nu(\pi)+1} = (-1)^{n+1} \frac{n}{n+1},$

where the sum is over all permutations of, is the sign of, i.e. if is even and if is odd, and is the number of fixed points of .

Now the sign of is given by

where the product is over all cycles c of, as explained e.g. on the page on even and odd permutations.

Hence we consider the combinatorial class

$\mathfrak{P}( - \mathcal{Z} + \mathcal{V} \mathcal{Z} + \mathfrak{C}_1(\mathcal{Z}) + \mathcal{U}\mathfrak{C}_2(\mathcal{Z}) + \mathcal{U}^2\mathfrak{C}_3(\mathcal{Z}) + \mathcal{U}^3\mathfrak{C}_4(\mathcal{Z}) + \cdots)$

where marks one minus the length of a contributing cycle, and marks fixed points. Translating to generating functions, we obtain

$g(z, u, v) = \exp\left( -z + vz + \sum_{k\ge 1} u^{k-1} \frac{z^k}{k} \right)$

$\exp\left( -z + vz + \frac{1}{u} \log\frac{1}{1-uz} \right) = \exp(-z + vz) \left(\frac{1}{1-uz} \right)^{1/u}.$

Now we have

$n! g(z, -1, v) = n! \exp(-z + vz) (1 + z) = \sum_{\pi\in S_n} \sigma(\pi) v^{\nu(\pi)}$

and hence the desired quantity is given by

$n! \int_0^1 g(z, -1, v) dv = \sum_{\pi\in S_n} \frac{\sigma(\pi)}{\nu(\pi)+1}.$

Doing the computation, we obtain

$\int_0^1 g(z, -1, v) dv = \exp(-z)(1+z) \left(\frac{1}{z}\exp(z) - \frac{1}{z}\right)$

$\left(\frac{1}{z} + 1\right) \left(1 - \exp(-z)\right) = \frac{1}{z} + 1 - \exp(-z) - \frac{1}{z} \exp(-z).$

Extracting coefficients, we find that the coefficient of is zero. The constant is one, which does not agree with the formula (should be zero). For positive, however, we obtain

$n! \left( - \exp(-z) - \frac{1}{z} \exp(-z) \right) = n! \left( - (-1)^n \frac{1}{n!} - (-1)^{n+1} \frac{1}{(n+1)!} \right)$

$(-1)^{n+1} \left( 1 - \frac{1}{n+1} \right) = (-1)^{n+1} \frac{n}{n+1},$

which is the desired result.

As an interesting aside, we observe that may be used to evaluate the following determinant of an matrix:

$d(n) = \det(A_n) = \begin{vmatrix} a && b && b && \cdots && b \\ b && a && b && \cdots && b \\ b && b && a && \cdots && b \\ \vdots && \vdots && \vdots && \ddots && \vdots \\ b && b && b && \cdots && a \end{vmatrix}.$

where . Recall the formula for the determinant:

Now the value of the product on the right for a permutation is, where f is the number of fixed points of . Hence

$d(n) = b^n n! g\left(z, -1, \frac{a}{b}\right)= b^n n! \exp \left(\frac{a-b}{b} z\right) (1+z)$

which yields

$b^n \left(\frac{a-b}{b}\right)^n + b^n n \left(\frac{a-b}{b}\right)^{n-1} = (a-b)^n + n b (a-b)^{n-1}$

and finally

Read more about this topic: Random Permutation Statistics

Famous quotes containing the words problem, putnam and/or competition:

“The general public is easy. You don’t have to answer to anyone; and as long as you follow the rules of your profession, you needn’t worry about the consequences. But the problem with the powerful and rich is that when they are sick, they really want their doctors to cure them.”
—Molière [Jean Baptiste Poquelin] (1622–1673)

“Cut the pie any way you like, “meanings” just ain’t in the head!”
—Hilary Putnam (b. 1926)

“Playing games with agreed upon rules helps children learn to live by rules, establish the delicate balance between competition and cooperation, between fair play and justice and exploitation and abuse of these for personal gain. It helps them learn to manage the warmth of winning and the hurt of losing; it helps them to believe that there will be another chance to win the next time.”
—James P. Comer (20th century)