Binomial Coefficient Properties

The binomial coefficient \binom{n}{k} (read 'n choose k') is defined for non-negative integers n and k with k \leq n as: \binom{n}{k} = \frac{n!}{k!(n-k)!}. For example, \binom{6}{2} = \frac{6!}{2! \cdot 4!} = \frac{720}{2 \cdot 24} = 15. Also, \binom{n}{0} = 1 and \binom{n}{n} = 1 for all n.

Binomial coefficients are symmetric: \binom{n}{k} = \binom{n}{n-k}. This makes sense combinatorially — choosing k items to include is the same as choosing n - k items to exclude. For example, \binom{8}{3} = \binom{8}{5} = 56. In Pascal's Triangle, each row reads the same forwards and backwards because of this property.

Pascal's Identity states that \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. This is the rule that builds each entry in Pascal's Triangle by adding the two entries directly above it. Proof: \frac{(n-1)!}{(k-1)!(n-k)!} + \frac{(n-1)!}{k!(n-1-k)!} = \frac{(n-1)!\cdot k + (n-1)!\cdot(n-k)}{k!(n-k)!} = \frac{(n-1)! \cdot n}{k!(n-k)!} = \binom{n}{k}.

Substituting a = 1 and b = 1 into the Binomial Theorem (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k gives: 2^n = \sum_{k=0}^{n}\binom{n}{k}. So the sum of all binomial coefficients in row n equals 2^n. For n = 4: \binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4} = 1+4+6+4+1 = 16 = 2^4.

Substituting a = 1 and b = -1 into the Binomial Theorem gives: 0 = \sum_{k=0}^{n}\binom{n}{k}(-1)^k = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots. This means the sum of even-positioned coefficients equals the sum of odd-positioned coefficients, each equal to 2^{n-1}. For n=4: (1+6+1) = (4+4) = 8 = 2^3.

Vandermonde's Identity states: \sum_{j=0}^{r}\binom{m}{j}\binom{n}{r-j} = \binom{m+n}{r}. This identity is proven by considering two groups of m and n people, and counting the ways to choose r people in total. A special case is \sum_{k=0}^{n}\binom{n}{k}^2 = \binom{2n}{n}, which gives the central binomial coefficient.

The Hockey Stick (or Christmas Stocking) Identity states: \sum_{i=r}^{n}\binom{i}{r} = \binom{n+1}{r+1}. For example, \binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2} = 1+3+6+10 = 20 = \binom{6}{3}. In Pascal's Triangle, if you trace a diagonal and then cut across, the sum along the diagonal equals the corner entry — hence the 'hockey stick' shape.

• n — Row index (dimensionless): A non-negative integer indicating which row of Pascal's Triangle (and which power in the binomial expansion) is being considered.
• k — Selection index (dimensionless): An integer running from 0 to n, indicating which term in the binomial expansion is selected.
• \binom{n}{k} — Binomial coefficient (dimensionless): The number of ways to choose k objects from n, equal to n! divided by k!(n-k)!. It is the k-th coefficient in the expansion of (a+b)^n.
• 2^n — Total sum (dimensionless): The result of the sum, obtained by substituting a = b = 1 in the Binomial Theorem (1+1)^n = 2^n.

Problem. Use binomial coefficient properties to evaluate: (a) \binom{10}{7}, and (b) show that \binom{7}{0}-\binom{7}{1}+\binom{7}{2}-\binom{7}{3}+\binom{7}{4}-\binom{7}{5}+\binom{7}{6}-\binom{7}{7} = 0.

Answer. \binom{10}{7} = 120; the alternating sum equals 0 by the Binomial Theorem with b = -1.

• The Binomial Theorem — Binomial coefficient properties arise directly from the Binomial Theorem — identities like the sum rule and alternating sum are proved by substituting specific values into the expansion (a+b)^n.
• Pascal's Triangle — Pascal's Triangle is the visual arrangement of all binomial coefficients, and Pascal's Identity is the rule that generates each new row from the previous one.
• Combinations and Permutations — The formula \binom{n}{k} = \frac{n!}{k!(n-k)!} is derived from counting combinations, making combinatorics the foundational language for understanding binomial coefficients.
• Binomial Distribution — In probability, binomial coefficients determine the probabilities in the binomial distribution: P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}. The sum identity confirms all probabilities sum to 1.