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,
(nk)−(n−3k)=(n−1k−1)+(n−2k−1)+(n−3k−1)
Answer
Repeatedly, use the identity (Pascal's Identity), namely
(nk)=(n−1k)+(n−1k−1).
Note that
((nk)−(n−1k−1))−(n−2k−1)−(n−3k−1)−(n−3k)
equals
(n−1k)−(n−2k−1)−(n−3k−1)−(n−3k)
which equals
(n−2k)−(n−3k−1)−(n−3k)
which equals
(n−3k)−(n−3k)=0
as desired.
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