Difference between revisions of "Binomial coefficient (n choose 0) equals 1"
From specialfunctionswiki
| Line 7: | Line 7: | ||
From the definition, | From the definition, | ||
$${n \choose k} = \dfrac{n!}{k! (n-k)!},$$ | $${n \choose k} = \dfrac{n!}{k! (n-k)!},$$ | ||
| − | so for $k=0$ we get, using the fact that $0!=1$, | + | so for $k=0$ we get, using the fact that [[0!=1|$0!=1$]], |
$${n \choose 0} = \dfrac{n!}{0! (n-0)!} = \dfrac{n!}{n!} = 1,$$ | $${n \choose 0} = \dfrac{n!}{0! (n-0)!} = \dfrac{n!}{n!} = 1,$$ | ||
as was to be shown. | as was to be shown. | ||
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
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $3.1.5$
