Difference between revisions of "Pascal's Identity"

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'''Pascal's Identity''' is a common and useful theorem in the realm of [[combinatorics]] dealing with [[combinations]] (also known as binomial coefficients), and is often used to reduce large, complicated [[combinations]].
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'''Pascal's identity''' is a common and useful theorem in the realm of [[combinatorics]] dealing with [[combinations]] (also known as binomial coefficients), and is often used to reduce large, complicated [[combinations]].
  
Pascal's Identity is also known as Pascal's Rule, Pascal's Formula, and occasionally Pascal's Theorem.
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Pascal's identity is also known as Pascal's rule, Pascal's formula, and occasionally Pascal's theorem.
  
 
== Theorem ==
 
== Theorem ==
Pascal's Identity states that
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Pascal's identity states that
  
 
<math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math>
 
<math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math>
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==History==
 
==History==
Pascal's Identity was probably first derived by [[Blaise Pascal]], a 19th century French mathematician, whom the theorem is named after. Pascal also did extensive other work on combinatorics, which is why [[Pascal's Triangle]] is named after him.
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Pascal's identity was probably first derived by [[Blaise Pascal]], a 19th century French mathematician, whom the theorem is named after. Pascal also did extensive other work on combinatorics, which is why [[Pascal's Triangle]] is named after him.
 
==See Also==
 
==See Also==
 
*[[Combinatorics]]
 
*[[Combinatorics]]

Revision as of 20:41, 8 December 2007

Pascal's identity is a common and useful theorem in the realm of combinatorics dealing with combinations (also known as binomial coefficients), and is often used to reduce large, complicated combinations.

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 $\{ k,n \in \bbfont{N} | k<n \}$ (Error compiling LaTeX. Unknown error_msg)

Proof

We have $\{ k,n \in \bbfont{N} | k<n \}$ (Error compiling LaTeX. Unknown error_msg):

$\binom{n-1}{k-1}+\binom{n-1}{k}=\frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{k!(n-k-1)!}$

$=(n-1)!\left[\frac{k}{k!(n-k)!}+\frac{n-k}{k!(n-k)!}$ (Error compiling LaTeX. Unknown error_msg)

$=(n-1)!\frac{n}{k!(n-k)!}$

$=\frac{n!}{k!(n-k)!}$

$=\binom{n}{k}$

Alternate Proof

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

\[{n \choose k}={n-1\choose k-1}+{n-1\choose k}\]

History

Pascal's identity was probably first derived by Blaise Pascal, a 19th century French mathematician, whom the theorem is named after. Pascal also did extensive other work on combinatorics, which is why Pascal's Triangle is named after him.

See Also

External Links