+102 Can somebody prove that \(\binom{n-1}{r-1}+\binom{n-1}{r} = \binom{n}{r}\)?
Thanks,
AnxiousLlama
+677 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
