Algebraic Proofs in Combinatorics $\binom{n + m}{2} = nm + \binom{n}{2} + \binom{m}{2}$
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How to algebraically prove $\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}$?
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Algebraic proof of $\sum_{i=0}^k{{n \choose i}{m \choose {k-i}}}= {{m+n}\choose k}$
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Give the combinatorial proof of the identity $\sum\limits_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$
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Show $\sum_{k=0}^{m} \binom{n-k}{m-k}=\binom{n+1}{m}$
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Binomial-coefficients if, k, m, n natural numbers and k \leq n the result of
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Combinatorics Identity about Catalan numbers: $\sum_{k=0}^n \frac{1}{k+1}\binom{2k}k \binom{2n-2k}{n-k}=\binom{2n+1}n$
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Prove the identity $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}=0$ using a combinatorial interpretation of the positive and negative terms
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Combinatorial proof that $\sum \limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$ when $n$ is even
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Prove $\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}$
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Prove ${n - 1 \choose k - 1} + {n - 2 \choose k - 1} + {n - 3 \choose k - 1} + \dots + {k - 1 \choose k - 1} = {n \choose k}$
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How can you prove $\binom {2n}{2} = 2\binom{n}{2} + n^2$?
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Revisited: Binomial Theorem: An Inductive Proof
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Prove that $\binom{m+n}{m}=\sum\limits_{i=0}^m \binom{m}{i}\binom{n}{i}$
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Combinatorial explanation for why $n^2 = {n \choose 2} + {n+1 \choose 2}$
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Algebraic proof for the identity $\sum_{k=0}^m(-1)^{m-k}{{n \choose {k}}}= {{n-1}\choose m}$
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With $a,\ b,\ m,\ n\in \mathbb{Z}$. Show that if ${\rm gcd}(m,n)=1$, then ${\rm gcd}(ma+nb,mn)={\rm gcd}(a,n){\rm gcd}(b,m)$
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Binomial-Theorem proof