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Can somebody prove that  \(\binom{n-1}{r-1}+\binom{n-1}{r} = \binom{n}{r}\)?

 

Thanks, 

AnxiousLlama

 Oct 11, 2021
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Hello AnxiousLlama,

 

I am problably too late and I'm not sure, but still I give a try:

 

\(\binom{n-1}{r-1}+\binom{n-1}{r}\) 

 

order the terms using the commutative law

 

\(\binom{n-1}{r}+\binom{n-1}{r-1}\)

 

simplify with the Pascal's triangle \(\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}\) to 

 

\(\binom{n}{r}\) so it's true.

 

Straight

 Oct 24, 2021

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