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For nonnegative integers $r,n$, the alternating binomial moment satisfies $S_r(n)=0$ when $r<n$ and $S_n(n)=(-1{)}^{n}n!$. Expand the polynomial $X^r$ in falling factorials and differentiate the binomial theorem to prove these identities. Equivalently, alternating binomial coefficients implement a finite difference that annihilates polynomials of degree less than $n$.

For an integer base $p\ge 2$, a nonnegative integer has a finite positional representation with digits $k_i\in \{0,\dots ,p-1\}$. Repeated Euclidean division gives the digits uniquely; appending zero high-place digits has no effect. The base need not be prime for uniqueness, but a prime base is useful in Lucas theorem and modular binomial coefficients.

The binomial coefficientcounts the $k$-element subsets of an $n$-element set. Equivalently, it counts strings containing $k$ copies of one symbol and $n-k$ copies of another.

For a positive integer $N$, all binomial coefficients $\left(\genfrac{}{}{0ex}{}{N}{j}\right)$ with $0<j<N$ are even exactly when $N$ is a power of two. Write the binary expansion $N={\sum }_{i\in I}{2}^i$. In the polynomial ring ${\mathbb{F}}_2[t]$, repeated squaring givesIf $I$ has one element, there are no interior terms. If it has more than one, the coefficient of $t^{2^{\min I}}$ is one and that exponent lies strictly between zero and $N$. This characterization gives a useful parity obstruction directly from the binary expansion.

For $n\ge 0$, define $B_0(X)=1$ and $B_n(X)=X(X-1)\cdots (X-n+1)/n!$. The binomial coefficient formula shows $B_n(m)\in \mathbb{Z}$ for every nonnegative integer $m$. Since these integers are dense in the p-adic integers, continuity of the polynomial implies $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$, so it is an integer-valued polynomial there and $\Vert B_n{\Vert }_{\infty }=1$. The Pascal's identity gives $\Delta B_{n+1}=B_n$ for the forward difference operator.

The central binomial coefficient satisfies $4^m/(2m+1)\le \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le 4^m$. Prime factors between $m$ and $2m$ give a dyadic upper bound for the Chebyshev theta function. The prime-power exponents in the factorial quotient are sums of zeros and ones, so the central coefficient gives a linear lower bound for the Second Chebyshev function. Higher prime powers contribute only $O(\sqrt{x}\log x)$; hence the prime-counting function has upper and lower bounds of order $x/\log x$.

Suppose a continuous map $g:S^n\times S^n\to S^n$ changes sign on negating either input. Its quotient $\bar{g}:{\mathbb{RP}}^n\times {\mathbb{RP}}^n\to {\mathbb{RP}}^n$ pulls back the real tautological line bundle to the tensor product of vector bundles $p_1^{\ast }{\gamma }_n\otimes p_2^{\ast }{\gamma }_n$. The fiber map sends $(sx)\otimes (ty)$ to $stg(x,y)$. The First Stiefel–Whitney class of a tensor product of real line bundles consequently gives ${\bar{g}}^{\ast }a=a_1+a_2$. The Künneth theorem and the mod-two cohomology ring of real projective space implyEvery interior binomial coefficient in row $n+1$ must be even. By binomial coefficients with even interior terms, $n=2^k-1$ for some $k\ge 0$. This is a necessary condition, not an assertion of existence in every such dimension.

Among $k$-Sperner subfamilies of $\mathcal{P}([n])$, the maximum cardinality is the sum of the $k$ largest binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. The union of the corresponding levels attains the bound. For the upper bound, the Lubell mass is at most $k$, and assigning its available mass to the levels with largest binomial coefficients maximizes cardinality.

The even part of a polynomial $P$ is $(P(z)+P(-z))/2$. It retains exactly the terms whose exponents are even. Applied to the binomial theorem, it extracts sums of even-indexed binomial coefficients.

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.

For a nonempty $r$-uniform set family, define $x>r-1$ by $|\mathcal{F}|=\left(\genfrac{}{}{0ex}{}{x}{r}\right)$ using real binomial coefficients. Then its lower shadow has at least $\left(\genfrac{}{}{0ex}{}{x}{r-1}\right)$ members. This continuous estimate is often easier to use than the exact integer expansion in the Kruskal-Katona theorem.

A short proof uses coordinate shifts of a set family. In a fully left-shifted set family split at coordinate one into ${\mathcal{F}}_0$ and ${\mathcal{F}}_1$, deleting one in the present section. The shifting property implies $\partial {\mathcal{F}}_0\subseteq {\mathcal{F}}_1$, and the total lower shadow has size $|{\mathcal{F}}_1|+|\partial {\mathcal{F}}_1|$. Induct on uniformity and, at fixed uniformity, on set family size. If $|{\mathcal{F}}_1|<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity and the size mathematical induction force $|\partial {\mathcal{F}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, a contradiction. The uniformity mathematical induction applied to ${\mathcal{F}}_1$, followed by the Pascal's identity, proves the estimate. Integer $x$ is sharp for the full $r$-level on $x$ coordinates.

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.

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.