Difference between revisions of "Binomial coefficient (n choose 0) equals 1"

From specialfunctionswiki
Jump to: navigation, search
(Created page with "==Theorem== The following formula holds: $${n \choose 0} = 1,$$ where ${n \choose 0}$ denotes the binomial coefficient. ==Proof== ==References== * {{BookReference|Handbo...")
 
 
(2 intermediate revisions by the same user not shown)
Line 5: Line 5:
  
 
==Proof==
 
==Proof==
 +
From the definition,
 +
$${n \choose k} = \dfrac{n!}{k! (n-k)!},$$
 +
so for $k=0$ we get, using the fact that [[0!=1|$0!=1$]],
 +
$${n \choose 0} = \dfrac{n!}{0! (n-0)!} = \dfrac{n!}{n!} = 1,$$
 +
as was to be shown.
  
 
==References==
 
==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Binomial coefficient ((n+1) choose k) equals (n choose k) + (n choose (k-1))|next=}: 3.1.5
+
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Binomial coefficient ((n+1) choose k) equals (n choose k) + (n choose (k-1))|next=Binomial coefficient (n choose n) equals 1}}: $3.1.5$
 +
 
 +
[[Category:Theorem]]
 +
[[Category:Proven]]

Latest revision as of 19:41, 9 October 2016

Theorem

The following formula holds: $${n \choose 0} = 1,$$ where ${n \choose 0}$ denotes the binomial coefficient.

Proof

From the definition, $${n \choose k} = \dfrac{n!}{k! (n-k)!},$$ so for $k=0$ we get, using the fact that $0!=1$, $${n \choose 0} = \dfrac{n!}{0! (n-0)!} = \dfrac{n!}{n!} = 1,$$ as was to be shown.

References