Moreover, for each positive integern, there exist connected graphs with **exactlyn**¡ 1 edges. Removing any edge from a connected graph of order n with **exactlyn**¡ 1 edges leavesa disconnected graph; hence each edge of a connected graph of ordernwith **exactlyn**¡ 1 edges is abridge.

THE SYMMETRIC GROUP HAROLD COOPER 1. Permutations Definition1.1. A permutation of a finite set Sis a bijection: S! S. Lemma1.1. There are **exactlyn**! permutations of ann-element set.

The most important is the result of performing **exactlyn** Bernoullitrials, each independent of the others and having the same probability of successp.

Fora set Sofsizen, there are **exactlyn**! permutations. To prove this, first observe that the number of permutations of a setS is finite and is dependent only on the cardinality ofS (andnotonany additional structure with which Smay be endowed).

An application of the Gram-Schmidt orthogonalisation process gives a sequencef j n: n2 N 0 gofreal-valued, linearly independent polynomials each in the Hilbertspace L 2 ( I ; µ ) and respectively of degree **exactlyn**, with the orthogonal property Z I j m (x) j n (x) d µ (x) =0, forallm, n2 N 0 withm6=n.

Show that anyn-goncanbe subdivided into **exactlyn** 2 triangles so that every triangle vertex is one of the original vertices of then-gon. Proof: This one starts atn=3; forn<3 it is true by default, since there is no such thing asa 2-gonor1-gon.

To avoid the special case of multiple roots, we can talk about "generic"behavior: almost all polynomials of degreenhave **exactlyn** complex roots. Here"almost all"means all but a set of measure zero, where the measure is defined by identifyinga polynomial with the point in Rn consisting of thetupleof ...

We also show that such an algebra is always contained in a commutative subalgebra of M n (k) of dimension **exactlyn**. 1. Introduction. Recall that if kisa eld, a matrix A 2M n (k) is said to be regular (or nonderogatory) if the minimal polynomial is equal toits characteristic polynomial, or equivalently, if V = k n ...

Show thatatreewithn vertices has **exactlyn** 1edges. Proof: We can prove this by induction. Ifn=1, the graph cannot have any edges or there would be a loop, with the vertex connecting to itself, so there must ben 1=0 edges.

If we consider the cube to be composed of the vertices and edges only, show that every n-cubehasa Hamiltonian circuit. 13. Show thatatreewithn vertices has **exactlyn** 1edges.