Pascal's rule

Not to be confused with Pascal's law.

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 ${n \choose k}$ is a binomial coefficient; one interpretation of which is 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 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 ${n \choose k}$ 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, choose X and k-1 elements from the remaining n-1 elements in the set. There are ${n-1 \choose k-1}$ such subsets.

To construct a subset of k elements not containing X, choose k elements from the remaining n-1 elements in the set. There are ${n-1 \choose k}$ 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, ${n-1 \choose k-1}+{n-1 \choose k}$.

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

Algebraic proof

Alternatively, the algebraic derivation of the binomial case follows.

${\begin{aligned}{n-1 \choose k}+{n-1 \choose k-1}&={\frac {(n-1)!}{k!(n-1-k)!}}+{\frac {(n-1)!}{(k-1)!(n-k)!}}\\&=(n-1)!\left[{\frac {n-k}{k!(n-k)!}}+{\frac {k}{k!(n-k)!}}\right]\\&=(n-1)!{\frac {n}{k!(n-k)!}}\\&={\frac {n!}{k!(n-k)!}}\\&={\binom {n}{k}}.\end{aligned}}$

Generalization

Pascal's rule can be generalized to multinomial coefficients. For any integer p such that $p\geq 2$, $k_{1},k_{2},k_{3},\dots ,k_{p}\in \mathbb {N} ^{*}$, and $n=k_{1}+k_{2}+k_{3}+\cdots +k_{p}\geq 1$,

${n-1 \choose k_{1}-1,k_{2},k_{3},\dots ,k_{p}}+{n-1 \choose k_{1},k_{2}-1,k_{3},\dots ,k_{p}}+\cdots +{n-1 \choose k_{1},k_{2},k_{3},\dots ,k_{p}-1}={n \choose k_{1},k_{2},k_{3},\dots ,k_{p}}$

where ${n \choose k_{1},k_{2},k_{3},\dots ,k_{p}}$ is the coefficient of the $x_{1}^{k_{1}}x_{2}^{k_{2}}\dots x_{p}^{k_{p}}$ 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 $p\geq 2$, $k_{1},k_{2},k_{3},\dots ,k_{p}\in \mathbb {N} ^{*}$, and $n=k_{1}+k_{2}+k_{3}+\cdots +k_{p}\geq 1$. Then

${\begin{aligned}&{}\quad {n-1 \choose k_{1}-1,k_{2},k_{3},\dots ,k_{p}}+{n-1 \choose k_{1},k_{2}-1,k_{3},\dots ,k_{p}}+\cdots +{n-1 \choose k_{1},k_{2},k_{3},\dots ,k_{p}-1}\\&={\frac {(n-1)!}{(k_{1}-1)!k_{2}!k_{3}!\cdots k_{p}!}}+{\frac {(n-1)!}{k_{1}!(k_{2}-1)!k_{3}!\cdots k_{p}!}}+\cdots +{\frac {(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots (k_{p}-1)!}}\\&={\frac {k_{1}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}+{\frac {k_{2}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}+\cdots +{\frac {k_{p}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}={\frac {(k_{1}+k_{2}+\cdots +k_{p})(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}\\&={\frac {n(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}={\frac {n!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}={n \choose k_{1},k_{2},k_{3},\dots ,k_{p}}.\end{aligned}}$

See also

• Pascal's triangle

Bibliography

• Merris, Russell. Combinatorics. John Wiley & Sons. 2003 ISBN 978-0-471-26296-1

External links

• "Central binomial coefficient". PlanetMath.
• "Binomial coefficient". PlanetMath.

This article incorporates material from Pascal's rule on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

This article incorporates material from Pascal's rule proof on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.