Pascal's Rule

Theorem

Let $\dbinom n k$ be a binomial coefficient.

For positive integers $n, k$ with $1 \le k \le n$:

$\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$

This is also valid for the real number definition:

$\forall r \in \R, k \in \Z: \dbinom r {k - 1} + \dbinom r k = \dbinom {r + 1} k$

Thus the binomial coefficients can be defined using the following recurrence relation:

$\dbinom n k = \begin{cases} 1 & : k = 0 \\ 0 & : k > n \\ \dbinom {n - 1} {k - 1} + \dbinom {n - 1} k & : \text{otherwise} \end{cases}$

Complex Numbers

For all $z, w \in \C$ such that it is not the case that $z$ is a negative integer and $w$ an integer:

$\dbinom z {w - 1} + \dbinom z w = \dbinom {z + 1} w$

where $\dbinom z w$ is a binomial coefficient.

Direct Proof

Let $n, k \in \N$ with $1 \le k \le n$.

$\ds \binom n k + \binom n {k - 1}$

$=$

$\ds \frac {n!} {k! \, \paren {n - k}!} + \frac {n!} {\paren {k - 1}! \, \paren {n - \paren {k - 1} }!}$

Definition of Binomial Coefficient

$\ds $

$=$

$\ds \frac {n! \, \paren {n - \paren {k - 1} } } {k! \, \paren {n - k}! \, \paren {n - \paren {k - 1} } } + \frac {n! \, k} {\paren {k - 1}! \, \paren {n - \paren {k - 1} }! \ k}$

multiplying by $1$

$\ds $

$=$

$\ds \frac {n! \, \paren {n - k + 1} } {k! \, \paren {n - k + 1}!} + \frac {n! \, k} {k! \, \paren {n - k + 1}!}$

Definition of Factorial

$\ds $

$=$

$\ds \frac {n! \, \paren {n - k + 1} + n! \, k} {k! \, \paren {n - k + 1}!}$

Addition of Fractions

$\ds $

$=$

$\ds \frac {n! \, \paren {n - k + 1 + k} } {k! \, \paren {n - k + 1}!}$

$\ds $

$=$

$\ds \frac {n! \, \paren {n + 1} } {k! \, \paren {n - k + 1}!}$

$\ds $

$=$

$\ds \frac {\paren {n + 1}!} {k! \, \paren {n + 1 - k}!}$

Definition of Factorial

$\ds $

$=$

$\ds \binom {n + 1} k$

Definition of Binomial Coefficient

$\blacksquare$

Combinatorial Proof

Suppose you were a member of a club with $n + 1$ members (including you).

Suppose it were time to elect a committee of $k$ members from that club.

From Cardinality of Set of Subsets, there are $\dbinom {n + 1} k$ ways to select the members to form this committee.

Now, you yourself may or may not be elected a member of this committee.

Suppose that, after the election, you are not a member of this committee.

Then, from Cardinality of Set of Subsets, there are $\dbinom n k$ ways to select the members to form such a committee.

Now suppose you are a member of the committee. Apart from you, there are $k - 1$ such members.

Again, from Cardinality of Set of Subsets, there are $\dbinom n {k - 1}$ ways of selecting the other $k - 1$ members so as to form such a committee.

In total, then, there are $\dbinom n k + \dbinom n {k - 1}$ possible committees.

Hence the result.

$\blacksquare$

Proof for Real Numbers

$\ds \paren {r + 1} \binom r {k - 1} + \paren {r + 1} \binom r k$

$=$

$\ds \paren {r + 1} \binom r {k - 1} + \paren {r + 1} \binom r {r - k}$

Symmetry Rule for Binomial Coefficients

$\ds $

$=$

$\ds k \binom {r + 1} k + \paren {r - k + 1} \binom {r + 1} {r - k + 1}$

Factors of Binomial Coefficient

$\ds $

$=$

$\ds k \binom {r + 1} k + \paren {r - k + 1} \binom {r + 1} k$

Symmetry Rule for Binomial Coefficients

$\ds $

$=$

$\ds \paren {r + 1} \binom {r + 1} k$

Dividing by $\paren {r + 1}$ yields the result.

$\blacksquare$

Also presented as

Some sources present as:

$\dbinom n k + \dbinom n {k + 1} = \dbinom {n + 1} {k + 1}$

Others present it in the form:

$\dbinom {n - 1} {k - 1} + \dbinom {n - 1} k = \dbinom n k$

Examples

$\binom 1 4$ plus $\binom 2 4$

$\ds \dbinom 4 1 + \dbinom 4 2$

$=$

$\ds 4 + 6$

$\ds $

$=$

$\ds 10$

$\ds $

$=$

$\ds \dbinom 5 2$

$\binom 1 5$ plus $\binom 2 5$

$\ds \dbinom 5 1 + \dbinom 5 2$

$=$

$\ds 5 + 10$

$\ds $

$=$

$\ds 15$

$\ds $

$=$

$\ds \dbinom 6 2$

$\binom 2 5$ plus $\binom 3 5$

$\ds \dbinom 5 2 + \dbinom 5 3$

$=$

$\ds 10 + 10$

$\ds $

$=$

$\ds 20$

$\ds $

$=$

$\ds \dbinom 6 3$

Also known as

Some sources refer to as Pascal's identity.

Also see

Definition:Pascal's Triangle

Source of Name

This entry was named for Blaise Pascal.

Sources

1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.4$: Binomial Coefficients
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis: Theorem $19.10$
1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.6$
1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.18$: Sequences Defined Inductively: Exercise $3 \ \text{(c)}$
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $12$
1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $35$
1992: Larry C. Andrews: Special Functions of Mathematics for Engineers (2nd ed.) ... (previous) ... (next): $\S 1.2.4$: Factorials and binomial coefficients: $1.30$
1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $35$
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.6.$
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): binomial coefficient $\text {(iii)}$
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Pascal's triangle
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): selection
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): binomial coefficient $\text {(iii)}$