Difference between revisions of "Pascal's Identity"

Latest revision as of 14:43, 11 April 2024

Pascal's Identity is a useful theorem of combinatorics dealing with combinations (also known as binomial coefficients). It can often be used to simplify complicated expressions involving binomial coefficients.

Pascal's Identity is also known as Pascal's Rule, Pascal's Formula, and occasionally Pascal's Theorem.

Theorem

Pascal's Identity states that

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

for any positive integers $k$ and $n$ . Here, $\binom{n}{k}$ is the binomial coefficient $\binom{n}{k} = {}_nC_k = C_k^n$ .

This result can be interpreted combinatorially as follows: the number of ways to choose $k$ things from $n$ things is equal to the number of ways to choose $k-1$ things from $n-1$ things added to the number of ways to choose $k$ things from $n-1$ things.

Proof

If $k > n$ then $\binom{n}{k} = 0 = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$ and so the result is trivial. So assume $k \leq n$ . Then

$\begin{eqnarray*}\binom{n-1}{k-1}+\binom{n-1}{k}&=&\frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{k!(n-k-1)!}\\ &=&(n-1)!\left(\frac{k}{k!(n-k)!}+\frac{n-k}{k!(n-k)!}\right)\\ &=&(n-1)!\cdot \frac{n}{k!(n-k)!}\\ &=&\frac{n!}{k!(n-k)!}\\ &=&\binom{n}{k}. \qquad\qquad\square\end{eqnarray*}$

Alternate Proofs

Here, we prove this using committee forming.

Consider picking one fixed object out of $n$ objects. Then, we can choose $k$ objects including that one in $\binom{n-1}{k-1}$ ways.

Because our final group of objects either contains the specified one or doesn't, we can choose the group in $\binom{n-1}{k-1}+\binom{n-1}{k}$ ways.

But we already know they can be picked in $\binom{n}{k}$ ways, so

${n \choose k}={n-1\choose k-1}+{n-1\choose k} \qquad \qquad \square$

Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term $\binom{n}{k}$ . Above that, we would see the terms ${n-1\choose k-1}$ and ${n-1\choose k}$ . Due to the definition of Pascal's Triangle, ${n \choose k}={n-1\choose k-1}+{n-1\choose k}$ .

History

Pascal's identity was probably first derived by Blaise Pascal, a 17th century French mathematician, whom the theorem is named after.

Pascal also did extensive other work on combinatorics, including work on Pascal's triangle, which bears his name. He discovered many patterns in this triangle, and it can be used to prove this identity. The method of proof using that is called block walking.

External Links

@@ Line 8: / Line 8: @@
 <math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math>
-for any [[positive integer]]s <math>k</math> and <math>n</math>.  Here, <math>\binom{n}{k}</math> is the binomial coefficient <math>\binom{n}{k} = nCk = C_k^n</math>.
+for any [[positive integer]]s <math>k</math> and <math>n</math>.  Here, <math>\binom{n}{k}</math> is the binomial coefficient <math>\binom{n}{k} = {}_nC_k = C_k^n</math>.
 This result can be interpreted combinatorially as follows: the number of ways to choose <math>k</math> things from <math>n</math> things is equal to the number of ways to choose <math>k-1</math> things from <math>n-1</math> things added to the number of ways to choose <math>k</math> things from <math>n-1</math> things.
@@ Line 44: / Line 44: @@
 *[[Combination]]
 *[[Committee forming]]
+*[[Combinatorial_identity#Hockey-Stick_Identity]]
 ==External Links==
-*[http://planetmath.org/encyclopedia/PascalsRule.html Pascal's Identity at Planet Math]
+*[https://planetmath.org/pascalsrule Pascal's Identity at Planet Math]
 *[http://mathworld.wolfram.com/PascalsFormula.html Pascal's Identity at Wolfram's Math World]
 [[Category:Combinatorics]]
 [[Category:Theorems]]

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