Binomial coefficient (n choose 0) equals 1

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: {{ #if: |{{{2}}}|Milton Abramowitz}}{{#if: Irene A. Stegun|{{#if: |, {{ #if: |{{{2}}}|Irene A. Stegun}}{{#if: |, [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]{{#if: |, [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]}}| and [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]}}| and {{ #if: |{{{2}}}|Irene A. Stegun}}}}|}}: [[Book:Milton Abramowitz/Handbook of mathematical functions{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|Handbook of mathematical functions{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}}|{{#if: | ({{{ed}}} ed.)}}}}]]{{#if: | (translated by [[Mathematician:{{{translated}}}|{{ #if: |{{{2}}}|{{{translated}}}}}]])}}{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: Binomial coefficient ((n+1) choose k) equals (n choose k) + (n choose (k-1)) | ... (previous)|}}{{#if: Binomial coefficient (n choose n) equals 1 | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}: $3.1.5$