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For nonnegative integers with $n_1+\cdots +n_r=n$, the multinomial coefficientcounts arrangements of a multiset containing $n_j$ copies of symbol $j$.

For occupations $W_n=Wp(n)$ of $W$ distinguishable copies, the number of arrangements is the multinomial coefficient $\mathcal{N}=W!/{\prod }_n{W}_n!$. The ensemble entropy is $S_W=k_B\log \mathcal{N}$. Stirling's formula and $\sum W_n=W$ giveDividing by the number of independent copies and taking the large-ensemble limit recovers $S=-k_B{\sum }_{n}p(n)\log p(n)$. Terms with zero occupation are interpreted by the usual $0\log 0=0$ convention.

Let $S_0=0$ and let each increment of the simple symmetric random walk be uniformly chosen from $\{\pm e_1,\pm e_2,\pm e_3\}$. A return requires an even number of steps. In $2n$ steps returning to zero, let $a,b,c$ be the respective numbers of positive steps in the three directions; there must also be $a,b,c$ negative steps, so $a+b+c=n$. Counting step sequences givesThe terms $p_{abc}=n!/(3^{n}a!b!c!)$ form a multinomial distribution, so $\sum p_{abc}^2\le (\max p_{abc})\sum p_{abc}=\max p_{abc}$. The permitted combinatorial inequalities give $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\le C_1{n}^{-1/2}$ and ${\max }_{a+b+c=n}{p}_{abc}\le C_2/n$ for $n\ge 1$. The second bound also follows by noting that the largest multinomial coefficient occurs when $a,b,c$ differ by at most one and then applying Stirling formula. ThusTo prove the origin is a transient state, let $q$ be the probability of at least one return after time zero. The Strong Markov property at successive returns gives $\mathbb{P}(N\ge k)=q^k$ for the number $N$ of later returns. Therefore $\mathbb{E}(N+1)={\sum }_j{u}_j$ would be infinite if $q=1$. The bound forces $q<1$, proving that the origin is a transient state. Translation invariance gives the same conclusion at every lattice point. The coordinate walks need not be treated as independent; the counting via multinomial coefficients already incorporates their dependence.

Let $D_n(a)$ denote the constant term, retaining the dimension as a subscript when it changes. We use the Good recurrence for the Dyson constant term. For algebraically independent variables, Lagrange interpolation polynomial applied to the constant polynomial one givesAt $t=0$, this becomes the rational identityWhen every $a_i>0$, multiplying by the given Laurent polynomial cancels one factor in row $i$ of the product and givesThis is a polynomial identity after cancellation; no choice of an expansion of a rational function is involved.

The boundary case is essential. If $a_i=0$, the factors in row $i$ are absent. The only factors involving $X_i$ are then $(1-X_j/X_i{)}^{a_j}$ for $j\ne i$, and all their powers of $X_i$ are nonpositive. To obtain power zero in $X_i$, one must select the constant term one from each of them. Taking the constant term in $X_i$ therefore deletes that variable and exponent:For $n=1$, the empty product is one, so $D_1(a_1)=1$. The all-zero exponent vector also gives one.

Now put $M_n(a)=m!/{\prod }_i{a}_i!$, where $m={\sum }_i{a}_i$. This multinomial coefficient obeys the same deletion rule for a zero exponent. For positive exponents,Induction on $n$, and within each dimension on $m$, consequently determines $D_n$ uniquely and identifies it with $M_n$. This proves the Dyson constant-term identity:

The mean extension in an ensemble of $A$ molecules iswith constraints ${\sum }_i{a}_i=A$ and ${\sum }_i{a}_i{l}_i=AL$. For occupancies $\{a_i\}$, the number of assignments of the $A$ distinguishable ensemble members to the states is the multinomial coefficientThus maximizing the probability at fixed $A$ and $L$ is equivalent to maximizing the stated Lagrange multiplier expression.

Because $A\gg 1$, the Stirling formula gives $\log (a_i!)\simeq a_i\log a_i-a_i$. Differentiating with respect to each $a_i$ givesNormalization by ${\sum }_i{a}_i=A$ therefore yields the probability mass functionHere $Z$ is the fixed-tension partition function that normalizes the probabilities.

Let $D(a)$ denote the stated constant term and suppose every $a_i>0$. The rational identitycan be verified after clearing denominators, or by Lagrange interpolation. Multiplying it by the Dyson product and taking constant terms gives the recursionThe multinomial coefficientobeys the same recursion by the multinomial form of Pascal's identity.

If $a_k=0$, taking the constant term in $X_k$ forces the zero term from every factor involving $X_k$ and reduces the expression to the $(n-1)$-variable Dyson product with $a_k$ omitted. The same boundary reduction holds for $M(a)$. Finally $D(0,\dots ,0)=M(0,\dots ,0)=1$. Induction on $n$ and on $a_1+\cdots +a_n$ therefore proves the Dyson constant-term identity

Under complete randomization, every assignment vector with exactly $n_z$ patients receiving treatment $z$ is equally likely. Thus, for $z=(z_1,\dots ,z_n)$ and $N_r(z)={\sum }_i{1}_{\{z_i=r\}}$,The reciprocal probability $n!/(n_0!n_1!n_2!)$ is the multinomial coefficient counting assignments with those arm sizes.