Pascal's rule

In mathematics, Pascal's rule (or Pascal's formula) is a combinatorial identity about binomial coefficients. It states that for positive natural numbers n and k, $${n-1\choose k} + {n-1\choose k-1} = {n\choose k},$$ where $$\tbinom{n}{k}$$ is a binomial coefficient; one interpretation of the coefficient of the $x^{k}$ term in the expansion of $(1 + x)^{n}$. There is no restriction on the relative sizes of $n$ and $k$, since, if $n < k$ the value of the binomial coefficient is zero and the identity remains valid.

Pascal's rule can also be viewed as a statement that the formula $$\frac{(x+y)!}{x!y!} = {x+y \choose x} = {x+y \choose y} $$ solves the linear two-dimensional difference equation $$ N_{x,y} = N_{x-1,y} + N_{x,y-1}, \quad N_{0,y} = N_{x,0} = 1 $$ over the natural numbers. Thus, Pascal's rule is also a statement about a formula for the numbers appearing in Pascal's triangle.

Pascal's rule can also be generalized to apply to multinomial coefficients.

Combinatorial proof
Pascal's rule has an intuitive combinatorial meaning, that is clearly expressed in this counting proof.

Proof. Recall that $$\tbinom{n}{k}$$ equals the number of subsets with k elements from a set with n elements. Suppose one particular element is uniquely labeled X in a set with n elements.

To construct a subset of k elements containing X, include X and choose k&thinsp;−&thinsp;1 elements from the remaining n&thinsp;−&thinsp;1 elements in the set. There are $$\tbinom{n-1}{k-1}$$ such subsets.

To construct a subset of k elements not containing X, choose k elements from the remaining n&thinsp;−&thinsp;1 elements in the set. There are $$\tbinom{n-1}{k}$$ such subsets.

Every subset of k elements either contains X or not. The total number of subsets with k elements in a set of n elements is the sum of the number of subsets containing X and the number of subsets that do not contain X, $$\tbinom{n-1}{k-1} + \tbinom{n-1}{k}$$.

This equals $$\tbinom{n}{k}$$; therefore, $$\tbinom{n}{k} = \tbinom{n-1}{k-1} + \tbinom{n-1}{k}$$.

Algebraic proof
Alternatively, the algebraic derivation of the binomial case follows. $$\begin{align} { n-1 \choose k } + { n-1 \choose k-1} & = \frac{(n-1)!}{k! (n - 1 - k)!} + \frac{(n-1)!}{(k - 1)!(n - k)!} \\ & = (n-1)! \left[ \frac{n - k}{k!(n - k)!} + \frac{k}{k!(n - k)!}\right] \\ & = (n-1)! \frac{n}{k!(n - k)!} \\ & = \frac{n!}{k!(n - k)!} \\ & = \binom{n}{k}. \end{align}$$

Generalization
Pascal's rule can be generalized to multinomial coefficients. For any integer p such that $$p \ge 2$$, $$k_1, k_2, k_3, \dots, k_p \in \mathbb{N}^{+}\!,$$ and $$n = k_1 + k_2 + k_3 + \cdots + k_p \ge 1$$, $${n-1\choose k_1-1,k_2,k_3, \dots, k_p}+{n-1 \choose k_1,k_2-1,k_3,\dots, k_p}+\cdots+{n-1\choose k_1,k_2,k_3,\dots,k_p-1} = {n\choose k_1, k_2, k_3, \dots, k_p}$$ where $${n\choose k_1, k_2, k_3, \dots, k_p}$$ is the coefficient of the $$x_1^{k_1}x_2^{k_2}\cdots x_p^{k_p}$$ term in the expansion of $$(x_1 + x_2 + \dots + x_p)^{n}$$.

The algebraic derivation for this general case is as follows. Let p be an integer such that $$p \ge 2$$, $$k_1, k_2, k_3, \dots, k_p \in \mathbb{N}^{+}\!,$$ and $$n = k_1 + k_2 + k_3 + \cdots + k_p \ge 1$$. Then $$ \begin{align} & {} \quad {n-1 \choose k_1-1,k_2,k_3, \dots, k_p}+{n-1 \choose k_1,k_2-1,k_3,\dots, k_p} + \cdots + {n-1 \choose k_1,k_2,k_3,\dots,k_p-1} \\ & = \frac{(n-1)!}{(k_1-1)!k_2!k_3! \cdots k_p!} + \frac{(n-1)!}{k_1!(k_2-1)!k_3!\cdots k_p!} + \cdots + \frac{(n-1)!}{k_1!k_2!k_3! \cdots (k_p-1)!} \\ & = \frac{k_1(n-1)!}{k_1!k_2!k_3! \cdots k_p!} + \frac{k_2(n-1)!}{k_1!k_2!k_3! \cdots k_p!} + \cdots + \frac{k_p(n-1)!}{k_1!k_2!k_3! \cdots k_p!} = \frac{(k_1+k_2+\cdots+k_p) (n-1)!}{k_1!k_2!k_3!\cdots k_p!} \\ & = \frac{n(n-1)!}{k_1!k_2!k_3! \cdots k_p!} = \frac{n!}{k_1!k_2!k_3! \cdots k_p!} = {n \choose k_1, k_2, k_3, \dots, k_p}. \end{align} $$