NISHIO Hirokazu[English][日本語]

binomial coefficient formula

  • eq1:

    • $\sum_{k=0}^n \binom{n}{k} = 2^n$
    • proof - binomial theorem
      • $(1 + 1) ^ n = \sum_{k=0}^n \binom{n}{k}$
      • $(1 + 1) ^ n = 2^n$
  • eq2:

    • $\sum_{k=0}^n (-1)^k \binom{n}{k} = 0$
    • proof - binomial theorem
      • $(-1 + 1) ^ n = \sum_{k=0}^n (-1)^k 1^{n-k} \binom{n}{k} = \sum_{k=0}^n (-1)^k \binom{n}{k}$
      • $(-1 + 1) ^ n = 0^n = 0$
  • eq3:

    • $\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} = 2^{n-1}$
    • proof
      • from eq1, eq2
      • $(1 + 1) ^ n + (-1 + 1) ^ n = 2^n$
      • $(1 + 1) ^ n + (-1 + 1) ^ n = \sum_{k=0}^n \binom{n}{k} + \sum_{k=0}^n (-1)^k \binom{n}{k} = 2 \sum_{i=0}^{\lfloor n/2 \rfloor} \binom{n}{2i}$
      • Since they cancel each other when k is odd
  • eq4: Field hockey stick identity

    • $\sum_{i=0}^k \binom{n+i}{i} = \binom{n+k+1}{k}$
  • eq5:

    • $\sum_{i=0}^k \sum_{j=0}^l \binom{i+j}{i} = \binom{k+l+2}{k+1}-1$
    • Pascal's triangulation
      • image
  • eq6: Vandermonde's identity

    • $\sum_{i=0}^k \binom{n}{i}\binom{m}{k - i} = \binom{n+m}{k}$
    • special case(k=n, m=n)
      • $\sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$
  • eq6-2 ARC110D

    • $\sum_i \binom{i}{A}\binom{N-i-1}{B} = \binom{N}{A+B+1}$
    • Ballistic Interpretation
      • image
  • eq6-3

  • eq7:

  • eq8:

ref - A collection of counting techniques


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