Pascal's rule

Template:Short description Script error: No such module "Distinguish". In mathematics, Pascal's rule (or Pascal's formula) is a combinatorial identity about binomial coefficients. The binomial coefficients are the numbers that appear in Pascal's triangle. Pascal's rule states that for positive integers n and k, $(\binom{n - 1}{k}) + (\binom{n - 1}{k - 1}) = (\binom{n}{k}),$ where $(\binom{n}{k})$ is the binomial coefficient, namely the coefficient of the Template:Math term in the expansion of Template:Math. There is no restriction on the relative sizes of Template:Mvar and Template:Mvar;^[1] in particular, the above identity remains valid when Template:Math since $(\binom{n}{k}) = 0$ whenever Template:Math.

Together with the boundary conditions $(\binom{n}{0}) = (\binom{n}{n}) = 1$ for all nonnegative integers n, Pascal's rule determines that $(\binom{n}{k}) = \frac{n!}{k! (n - k)!},$ for all integers Template:Math. In this sense, Pascal's rule is the recurrence relation that defines the binomial coefficients.

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

Combinatorial proof

File:Pascal's rule 4c1 plus 4c2 equals 5c2.svg

Illustrates combinatorial proof:

(\binom{4}{1}) + (\binom{4}{2}) = (\binom{5}{2}) .

Pascal's rule has an intuitive combinatorial meaning, that is clearly expressed in this counting proof.^[2]Template:Rp

Proof. Recall that $(\binom{n}{k})$ 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 − 1 elements from the remaining n − 1 elements in the set. There are $(\binom{n - 1}{k - 1})$ 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 $(\binom{n - 1}{k})$ 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, $(\binom{n - 1}{k - 1}) + (\binom{n - 1}{k})$ .

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

Algebraic proof

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

An alternative algebraic proof using the alternative definition of binomial coefficients: $(\binom{n}{k}) = \frac{n (n - 1) \dots (n - k + 1)}{k!}$ . Indeed

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

Since $(\binom{z}{k}) = \frac{z (z - 1) \dots (z - k + 1)}{k!}$ is used as the extended definition of the binomial coefficient when z is a complex number, thus the above alternative algebraic proof shows that Pascal's rule holds more generally when n is replaced by any complex number.

Generalization

Pascal's rule can be generalized to multinomial coefficients.^[2]Template:Rp For any integer p such that $p \geq 2$ , $k_{1}, k_{2}, k_{3}, \dots, k_{p} \in ℕ^{+},$ and $n = k_{1} + k_{2} + k_{3} + \dots + k_{p} \geq 1$ , $(\binom{n - 1}{k_{1} - 1, k_{2}, k_{3}, \dots, k_{p}}) + (\binom{n - 1}{k_{1}, k_{2} - 1, k_{3}, \dots, k_{p}}) + \dots + (\binom{n - 1}{k_{1}, k_{2}, k_{3}, \dots, k_{p} - 1}) = (\binom{n}{k_{1}, k_{2}, k_{3}, \dots, k_{p}})$ where $(\binom{n}{k_{1}, k_{2}, k_{3}, \dots, k_{p}})$ 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.^[2]Template:Rp Let p be an integer such that $p \geq 2$ , $k_{1}, k_{2}, k_{3}, \dots, k_{p} \in ℕ^{+},$ and $n = k_{1} + k_{2} + k_{3} + \dots + k_{p} \geq 1$ . Then $\begin{aligned} (\binom{n - 1}{k_{1} - 1, k_{2}, k_{3}, \dots, k_{p}}) + (\binom{n - 1}{k_{1}, k_{2} - 1, k_{3}, \dots, k_{p}}) + \dots + (\binom{n - 1}{k_{1}, k_{2}, k_{3}, \dots, k_{p} - 1}) \\ = \frac{(n - 1)!}{(k_{1} - 1)! k_{2}! k_{3}! \dots k_{p}!} + \frac{(n - 1)!}{k_{1}! (k_{2} - 1)! k_{3}! \dots k_{p}!} + \dots + \frac{(n - 1)!}{k_{1}! k_{2}! k_{3}! \dots (k_{p} - 1)!} \\ = \frac{k_{1} (n - 1)!}{k_{1}! k_{2}! k_{3}! \dots k_{p}!} + \frac{k_{2} (n - 1)!}{k_{1}! k_{2}! k_{3}! \dots k_{p}!} + \dots + \frac{k_{p} (n - 1)!}{k_{1}! k_{2}! k_{3}! \dots k_{p}!} = \frac{(k_{1} + k_{2} + \dots + k_{p}) (n - 1)!}{k_{1}! k_{2}! k_{3}! \dots k_{p}!} \\ = \frac{n (n - 1)!}{k_{1}! k_{2}! k_{3}! \dots k_{p}!} = \frac{n!}{k_{1}! k_{2}! k_{3}! \dots k_{p}!} = (\binom{n}{k_{1}, k_{2}, k_{3}, \dots, k_{p}}) . \end{aligned}$

References

Template:Reflist

Bibliography

Merris, Russell. Combinatorics. John Wiley & Sons. 2003 Template:ISBN

External links

This article incorporates material from Pascal's triangle 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.

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Pascal's rule

Contents

Combinatorial proof

Algebraic proof

Generalization

See also

References

Bibliography

External links

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Pascal's rule

Combinatorial proof

Algebraic proof

Generalization

See also

References

Bibliography

External links

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