# Exactlyn

### Trees

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

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 Bernoulli Distribution

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

### The story of the symmetric group

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).

### The (Bessel, Jacobi, Laguerre, Legendre)-type linear fourth ...

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.

### Induction Problems

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.

### Homology Notes 3 Degreeofamapf: Xn→Sn

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 ...

### COMMUTING PAIRS IN THE CENTRALIZERS OF 2-REGULAR MATRICES

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 ...

### Graph TheoryProblems and Solutions

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.

### Graph TheoryProblems

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.