For a prime number $p$, the binomial coefficient modulo $p$ factors into the corresponding digit binomial coefficients in the base-p expansions of the nonnegative integers $n,k$. Pad both expansions to the same length and use $\left(\genfrac{}{}{0ex}{}{a}{b}\right)=0$ when $b>a$. The identity follows by factoring $(1+x{)}^n$ into digit powers and using the Frobenius endomorphism; digit uniqueness identifies each coefficient.

For nonnegative integers $n,m$, the binomial coefficients satisfyThe identity follows by mathematical induction: adding the next term combines two adjacent binomial coefficients through Pascal's identity. An equivalent form is ${\sum }_{r=j}^N\left(\genfrac{}{}{0ex}{}{r}{j}\right)=\left(\genfrac{}{}{0ex}{}{N+1}{j+1}\right)$, a diagonal sum in Pascal's triangle. It provides closed forms for sums of binomial coefficients without expanding individual factorials.

Row $n$ of Pascal's triangle consists of the binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{0}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{n}\right)$, beginning with row $0$. Boundary entries are one, and Pascal's identity says each interior entry is the sum of the two entries above it. The hockey-stick identity evaluates sums along its diagonals.

The highlighted entry in the binomial coefficient illustration is $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. Split two-person teams according to whether they contain the fifth person: six choose both members from the other four, and four contain the fifth person together with one of those four. Thus $10=6+4$. All entries can be generated by the same local rule, so a large table of counts needs no repeated enumeration of its subsets.

Reading row seven. Label the entries in each row from $k=0$ on the left through $k=n$ on the right. Rows zero through seven are shown below. The highlighted entry $T_{7,3}=35$ comes from $T_{6,2}=15$ and $T_{6,3}=20$. Starting with row zero $(1)$, the edges remain one and Pascal's rule determines every interior entry, so mathematical induction uniquely determines the whole array.

The triangular and rectangular layouts describe the same entries. For a north-east lattice path ending at $(a,b)$, set $n=a+b$ and $k=b$, so $P(a,b)=T_{n,k}$. The predecessor $(a-1,b)$ has indices $(n-1,k)$, while $(a,b-1)$ has indices $(n-1,k-1)$. Thus the grid's west and south arrivals are exactly the triangle's two parents. This change of indices explains why the north-east lattice path recurrence and Pascal's rule agree.

There is also a global check on the local construction: row $n$ sums to $2^n$. Every subset of an $n$-element set appears exactly once among the classes of sizes $0,\dots ,n$, and each element has two inclusion choices by the rule of product. In row seven, $1+7+21+35+35+21+7+1=128=2^7$. Replacing every subset by its complement reverses the row, explaining $T_{n,k}=T_{n,n-k}$. The binomial theorem packages the whole row as the coefficients of $(1+x{)}^n$.

Blaise Pascal's Traité du triangle arithmétique, published in 1665, presents an arithmetic array and connects it to combinations, games of chance, and powers of binomial expressions. This makes the connection between a counting table and several applications visible in the original work. ^[ref] The University of Cambridge's account of the treatise explains its 1654 composition and the organization of these applications. ^[ref]

A self-delimiting code permits a decoder to identify the end of a code without an external length marker. Encoding a binary word of length $\ell$ by $\ell$ ones, a zero and the word uses $2\ell +1$ bits. This yields an $O(\log n)$ description of a natural number in an incompressibility proof.

A codimension-two bifurcation requires two independent parameter conditions to hold simultaneously. Simultaneous stationary and Hopf bifurcation thresholds require both kinds of critical mode in the reduced amplitude equation. A crossing of modes at different wavenumbers is different from a zero-frequency double-zero degeneration of one fixed-wave-number cubic.

A supercritical branch emerges on the parameter side where the original state loses stability. In the cubic Landau amplitude equation $\dot{A}=\mu A-g|A{|}^{2}A$ with $g>0$, the nonzero branch $|A{|}^2=\mu /g$ lies at $\mu >0$ and is stable to radial amplitude perturbations. Continuous phase symmetry still leaves a neutral phase direction. The direction of a branch and its stability should both be checked in a general problem of bifurcation theory.

A subcritical branch emerges on the parameter side where the original state remains stable. In the cubic Landau amplitude equation $\dot{A}=\mu A-g|A{|}^{2}A$ with $g<0$, the small nonzero branch lies at $\mu <0$ and is unstable to radial amplitude perturbations. Higher-order saturation can generate a finite-amplitude stable branch and hysteresis, but this does not follow from a cubic truncation alone.

The two Hankel functions $H_{\nu }^{(1)}=J_{\nu }+i{Y}_{\nu }$ and $H_{\nu }^{(2)}=J_{\nu }-i{Y}_{\nu }$ solve the Bessel differential equation. Their large-argument forms distinguish outgoing and incoming oscillations and have sector-dependent asymptotic expansions.

A multiplier $B(x,y)$ whose rescaled planar vector field has divergence of one strict sign can exclude periodic orbits in a simply connected region. For a system $\dot{x}=xF$, $\dot{y}=yG$ in the positive quadrant, $B=1/(xy)$ is often effective.

For real $-1\le \alpha \le 1$, this two-qubit density operator has equal $00$ and $11$ populations and off-diagonal coherences $\alpha /2$. It is a pure state exactly when $\alpha =\pm 1$, and a separable quantum state exactly when $\alpha =0$. Its partial transpose has eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$, directly witnessing entanglement for every nonzero admissible coherence.

For iid $N(\mu ,v)$ observations, the prior density $1/v$ relative to $d\mu dv$ gives posterior kernel $v^{-n/2-1}\exp [-\{S+n(\mu -\bar{x}{)}^2\}/(2v)]$. It is proper exactly when $n>1$ and $S=\sum (x_i-\bar{x}{)}^2>0$. The conditionals are $\mu \mid v,x\sim N(\bar{x},v/n)$ and $v\mid \mu ,x\sim \mathrm{IG}(n/2,{\sum }_i(x_i-\mu {)}^2/2)$. Integrating out $\mu$ changes the inverse-gamma shape to $(n-1)/2$.

The ${\Lambda }^0$ in the baryon octet is an isospin singlet with strangeness $-1$ and zero electric charge. Its ordinary valence quark content is $uds$.

The baryon octet contains an isospin triplet ${\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-}$. Each has strangeness $-1$; their valence quark contents are $uus$, $uds$, $dds$.

The baryon octet contains the isospin doublet ${\Xi }^0,{\Xi }^{-}$. Each has strangeness $-2$, with valence quark content $uss$, $dss$.

The invariant colour state of three fundamental quarks is the normalized completely antisymmetric tensor ${ϵ}_{abc}/\sqrt{6}$. With this colour factor, the Pauli exclusion principle requires the remaining spatial-spin-flavour wavefunction to be symmetric. Colour and approximate flavour symmetry are distinct SU(3) actions.

In a three-quark colour singlet, a symmetric spatial state requires symmetric spin-flavour symmetry. Completely symmetric decuplet flavour couples to symmetric spin $3/2$; mixed octet flavour couples to mixed spin $1/2$ in the symmetric permutation combination. Antisymmetric singlet flavour cannot have a symmetric S-wave orbital state, since ${\bigwedge }^3{\mathbb{C}}^2=0$ forbids a completely antisymmetric state of three spin-$1/2$ quarks. Mixed orbital and spin states can instead realize the required antisymmetry.

A baryon with constituent content four quarks and one antiquark, whose net baryon number is one. Such content admits more flavour representations than three quarks alone. A representation in the tensor product is not evidence by itself of a bound state.

For a barotropic fluid, an energy per mass $e$ can be defined by $de/d\rho =p/{\rho }^2$. For the isothermal equation of state, $e=c_s^2\ln (\rho /{\rho }_0)$. Adding kinetic, gravitational and magnetic energy gives the ideal magnetic energy flux $\mathbf{u}(E+p+B^2/(2{\mu }_0))-\mathbf{B}(\mathbf{u}\cdot \mathbf{B})/{\mu }_0$. This is a mechanical barotropic energy identity; it does not say that an isothermal gas is thermally isolated.

For a barotropic fluid with pressure $P=P(\rho )$, choose an energy density $U$ satisfying $\rho U^{\prime }(\rho )-U=P$. The continuity equation then gives ${\partial }_{t}U+\nabla \cdot (U\mathbf{u})=-P\nabla \cdot \mathbf{u}$. For an isothermal equation of state $P=c_s^2\rho$, one choice is $U=c_s^2\rho \ln (\rho /{\rho }_{\mathrm{ref}})$. A fixed reference density changes $U$ only by a conserved mass term. The same construction holds with surface density and vertically integrated pressure. This closure energy need not equal the microscopic internal energy of a gas maintained at fixed temperature.

For angular velocity depending only on cylindrical radius, the centrifugal acceleration is a gradient. Steady barotropic balance is $\nabla p/\rho =-\nabla \Psi$, hence specific enthalpy satisfies $h(\rho )+\Psi =\text{constant}$. On an invertible equation-of-state branch, mass density and pressure are functions of $\Psi$. Cylindrical rotation is essential to this scalar centrifugal-potential representation.