Saturday, 28 June 2014

combinatorics - Prove the following by two different methods, one combinatorial and one algebraic



Reading through my textbook I came across the following problem, and I am looking for some help solving it. I am asked to prove the following by two different methods, one combinatorial and one algebraic. If I could get help with either or both it would be great, thanks!



Prove that this identity is true,



\binom{n}{k} -\binom{n-3}{k} =\binom{n-1}{k-1} + \binom{n-2}{k-1} + \binom{n-3}{k-1}


Answer



Repeatedly, use the identity (Pascal's Identity), namely

\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}.
Note that
\left(\binom{n}{k}-\binom{n-1}{k-1}\right)-\binom{n-2}{k-1}-\binom{n-3}{k-1}-\binom{n-3}{k}
equals
\binom{n-1}{k}-\binom{n-2}{k-1}-\binom{n-3}{k-1}-\binom{n-3}{k}
which equals
\binom{n-2}{k}-\binom{n-3}{k-1}-\binom{n-3}{k}
which equals
\binom{n-3}{k}-\binom{n-3}{k}=0
as desired.



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