A layered solid form of carbon, with covalent bonds joining chemical atoms within each layer.

A bond dipole represents the separation of positive and negative electric charge associated with a chemical bond. In a neutral two-point charge model put $+q$ at ${\mathbf{r}}_A$ and $-q$ at ${\mathbf{r}}_B$, with $q>0$. Its physical electric dipole moment isIf $B$ is the negative end, the physical electric dipole moment points from $B$ to $A$. The customary crossed-tail chemical arrow points towards the negative, more electronegative end, from $A$ to $B$. It therefore has the opposite direction. State the convention before doing vector addition; use physical electric dipole moments in the interaction potential energy $U=-\mathbf{p}\cdot \mathbf{E}$, where $\mathbf{E}$ is the electric field.^[ref]

For a fixed neutral molecule, its complete electric dipole moment follows from all atomic nuclei and the electron density, not just point charges assigned to sites. If $n(\mathbf{r})$ is the expected number of electrons per unit volume, the corresponding charge density is ${\rho }_e(\mathbf{r})=-en(\mathbf{r})$; the nuclear terms contribute separately. Partitioning that distribution into bond dipoles is a model. Lone pairs and electric polarization within a chemical atom can contribute as well. A measured molecular electric dipole moment cannot uniquely recover every individual bond dipole or partial atomic charge.

Suppose two independent random variables have the same categorical distribution, with label probabilities $p_1,\dots ,p_m$. The disjoint equal-label events giveFor the original chosen probabilities $(1/2,1/4,1/8,1/8)$, this is $11/32=0.34375$, rather than the uniform four-label value $1/4$. The Cauchy-Schwarz inequality gives $1=({\sum }_i{p}_i{)}^2\le m{\sum }_i{p}_i^2$, so uniform probabilities minimize this two-draw collision probability. This pair formula alone does not supply the entire group collision probability for a nonuniform model.

The union bound gives $P_{n,m}\le \min (1,n(n-1)/(2m))$. Using $1-x\le e^{-x}$ on the distinct-label product givesThe lower bound reaches at least one half when $n(n-1)/(2m)\ge \log 2$. The simple approximation $n\approx \sqrt{2m\log 2}$ predicts a group size near 23 for $m=365$. It is the quadratic number of pairs that makes the square root appear.

For a precise limiting statement, let $m\to \infty$ and $n/\sqrt{m}\to c\ge 0$. Expanding the logarithm of the complement yieldsOn this scale the remainder tends to zero. Thus a fixed substantial collision probability appears after order $\sqrt{m}$ draws. For $m=2^b$ uniform output labels, the corresponding scale is order $2^{b/2}$; matching one specified birthday instead involves a number of draws of order $m$.

The birthday collision count counts equal-label unordered pairs. Write it as a sum of indicator random variables:This uses linearity of expectation; mutual independence of the pair indicators is unnecessary. For 23 uniform calendar labels, $\mathbb{E}[C]=253/365\approx 0.69315$, whereas $\mathbb{P}(C\ge 1)\approx 0.50730$. A configuration with three equal labels contributes three pairs, so counting pairs and recording whether any pair exists are different random variables.

Treating every pair event as mutually independent would give the wrong exact formula. For three draws, $X_1=X_2$ and $X_1=X_3$ force $X_2=X_3$. The probability that all three equalities hold is $1/m^2$, while the product of their individual probabilities is $1/m^3$ for $m>1$. This distinguishes independence of the original label draws from mutual independence of events built from overlapping draws.

In a group of total size $n\ge 1$, designate one member and ask whether any of the other $n-1$ labels equals that member's label. Conditional on the designated label, all the other draws avoid it with probability $1-1/m$. Independence givesFor $n=23$, this is approximately $5.86\%$, while the probability of some matching pair is approximately $50.73\%$. The designated-member question examines $22$ pairs; the any-pair question examines all $253$. A repeated label elsewhere in the group does not require a match with its designated member.

Every nonempty finite binary tree with $n$ vertices has exactly $n+1$ empty binary tree child slots. There are $2n$ slots altogether, and precisely $n-1$ occupied slots, one for each edge of its underlying tree. ThereforeThis counts missing child references, not vertices or leaves. In the mixed four-vertex example, $A$ has no empty binary tree child slots, $B$ has one, and $C,D$ each have two: $0+1+2+2=5$. The four-vertex chain also has five: one at each of its first three vertices and two at the last. For $n=0$, there are no vertices and hence no child fields; the empty binary tree instead has a single empty root position if it is represented by a root reference.

A second proof uses the recursive decomposition. Count every empty subtree occurrence, including the root position for the empty binary tree: let $e(\varnothing )=1$. For a nonempty binary tree, $e(T)=e(L)+e(R)$. If the two subtrees have $n_L,n_R$ vertices, structural induction gives $e(T)=(n_L+1)+(n_R+1)=n+1$. Distinct empty positions may share the same null value; these are occurrences, not separately allocated objects. This count is useful when analysing the base cases reached by a structural recursion that calls both subtrees of every vertex.

An atom interferometer supplies atomic preparation, coherent splitting and recombination, and a readout of the output populations. In a light-pulse apparatus, a cooled atomic cloud interacts with controlled laser pulses during flight, and state-selective detection measures the populations after the final pulse.

For ideal equal interfering amplitudes, one output has probability $P=[1+\mathcal{V}\cos \Delta \varphi ]/2$, where $\Delta \varphi$ is the relative wave phase and $\mathcal{V}$ the interferometric visibility. Scanning a laser phase produces fringes. Vibrations, phase noise, unequal path populations and information distinguishing the paths can reduce their visibility; simply detecting two atomic groups does not establish wave interference.

The house mouse, Mus musculus, is a small rodent associated with human environments and widely used in experimental biology. Laboratory strains provide controlled inherited backgrounds, while wild populations reveal variation shaped by ecology and ancestry.

A protractor is a graduated instrument for measuring or drawing planar angles, commonly a semicircular scale from $0^{\circ }$ to $180^{\circ }$. Place its centre at the vertex, align its zero baseline with one ray, and read where the other ray crosses the scale. A $60^{\circ }$ reading represents $\pi /3$ radians. A device with oppositely directed scales requires choosing the scale whose zero matches the aligned baseline.

An angle bisector is a ray dividing an angle into two equal angles. For nonopposite unit vectors $u,v$, its internal direction is $u+v$: the dot products with $u$ and $v$ are equal. The direction $u-v$ bisects the external angle. The three internal angle bisectors of a triangle meet at the incenter, equidistant from its side lines. The angle bisector theorem relates each bisector to side-length ratios.