Difference between revisions of "Binomial coefficient (n choose 0) equals 1"
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==Proof== | ==Proof== | ||
| + | From the definition, | ||
| + | $${n \choose k} = \dfrac{n!}{k! (n-k)!},$$ | ||
| + | so for $k=0$ we get, using the fact that $0!=1$, | ||
| + | $${n \choose 0} = \dfrac{n!}{0! (n-0)!} = \dfrac{n!}{n!} = 1,$$ | ||
| + | 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=Binomial coefficient (n choose n) equals 1}}: 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]] | ||
Revision as of 19:40, 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): $3.1.5$
