What are permutation statistics? Let’s write [n] for the set {1, 2, 3, …, n}. A permutation of size n is a bijection  :[n]  [n]. We usually write permutations as sequences, and denote their entries using subscripts. For example,  = 2413 is a permutation of size 4, and its last entry is  4 = 3. An inversion in  is a pair of entries (not necessarily consecutive) that appear in decreasing order; that is, i  j. If the entries are consecutive it is also called a descent. The permutation 2413 has two inversions (41 and 43) but it only has one descent (41). Let: A ( n, k ) = # of permutations of size n with k descents B ( n, k ) = # of permutations of size n with k inversions The inversion number and the descent number are examples of permutation statistics. The functions A(n,k) and B(n,k) are called the distributions of these statistics. Factorial Functions A factorial function of size n is a function from [n] to [n] that satisfies a i 2). Every factorial function has a 1 =1. The image of a is the set of distinct values of a: Im a = { j | for some i, a i = j }. The image size, |Im a|, is always an integer from 1 to n. It turns out that the image size is an Eulerian statistic. Theorem 1. The number of factorial functions of size n with |Im a| = k is exactly A(n,k). Proof: For each a, construct pi as follows. Start with an empty sequence for pi. Then, for each i in order from 1 to n, insert the value i after the value a i. BUT if that would cause i to be at the end of the sequence, put it at the start instead. AND if a i =i, put the value i at the end. For example, if a = 1132 we get pi = Note that if i is in the image of a, then the value i does NOT begin a descent. Well, this proof is failing. I need to have defined A(n,k) as the number of permutations with n-k descents, not k descents. That makes the definition ugly. How can I fix this? Theorem: The excedence number is Eulerian. Proof: We construct a different bijection from the factorial functions to permutations. Given a, construct pi as follows: For each i, insert the value i at position a i. For example, 1132 becomes Well, that’s no good. Maybe it will be the proof of the next theorem, but it isn’t the proof of this one. Bring it home Well, it looks as if there isn’t room for what I want to do. The SUM of a factorial functions is the sum of its values (minus n). Theorem: SUM is Mahonian, because SUM has the same distribution as INV. Theorem: MAJ is Mahonian, because MAJ has the same distribution as SUM. Therefore, THEOREM: MAJ and INV have the same distribution. The point of this poster is that the proof of the THEOREM, using factorial functions, is MUCH EASIER than any of the published proofs that work directly with permutations. But the poster doesn’t convey this point very well, because you can’t see the proofs of the theorems in this box, and the proofs of the theorems in the other block are garbled beyond recognition. But I know the proofs. And they’re very easy. REALLY. Factorial functions and permutation statistics Walter Stromquist Swarthmore College, Department of Mathematics & Statistics References Bender, D.J., E.M Bayne, and R.M. Brigham Lunar condition influences coyote (Canis latrans) howling. American Midland Naturalist 136: Brooks, L.D The evolution of recombination rates. Pages in The Evolution of Sex, edited by R.E. Michod and B.R. Levin. Sinauer, Sunderland, MA. Scott, E.C Evolution vs. Creationism: an Introduction. University of California Press, Berkeley. Society for the Study of Evolution Statement on teaching evolution.. Accessed 2005 Aug 9. For more information --- wait and see. There are lots of theorems and formulas about A(n,k) and B(n,k). Here are some tables of values: Acknowledgements This work (but not garbles) results from collaboration with Alex Burstein, David Chudzicki, Mike Stone, and maybe Sergey Kitaev. Different statistics can have identical distributions. A weak excedence in a permutation  is an entry that is at least as large as its position value --- that is, an instance of  i  i. For example, 2413 has two excedences: in position 1 (because 2  1) and in position 2 (because 4  2). The weak excedence number is a very different statistic from the descent number. For example, we have seen that 2413 has descent number 1, but weak excedence number 2. But the number of permutations of size n with k weak excedences is exactly A(n,k) --- that is, the weak excedence number has the same distribution as the descent number. (We’ll prove that in a moment.) Any statistic that has the distribution A(n,k) is called an Eulerian statistic. McMahon found a statistic that has the same distribution as the inversion number. It’s called the Major index. To compute it, note the starting positions of all the descents in . The Major index is the sum of all the starting positions. For example: 2413 has just one descent, 41, which starts in position 2. So, we add 2 (and nothing else) to get the Major index of 2413, which is 2. As a bigger example: has 4 descents , 32, 21, 65 starting at positions 1, 2, 3, 5. So its Major index is = 11. The Major index doesn’t seem to have much to do with the inversion number. But according to McMahon, they have the same distribution. The number of permutations of size n with Major index k is exactly B(n,k). Any statistic that has the distribution B(n,k) is called a Mahonian statistic A(n,k) We’ll put a table of B(n,k) here, too.