Documentation

Mathlib.Data.Nat.Choose.Basic

Binomial coefficients #

This file defines binomial coefficients and proves simple lemmas (i.e. those not requiring more imports). For the lemma that n.choose k counts the k-element-subsets of an n-element set, see Fintype.card_powersetCard in Mathlib/Data/Finset/Powerset.lean.

Main definition and results #

Tags #

binomial coefficient, combination, multicombination, stars and bars

def Nat.choose :

choose n k is the number of k-element subsets in an n-element set. Also known as binomial coefficients. For the fact that this is the number of k-element-subsets of an n-element set, see Fintype.card_powersetCard.

Equations
    Instances For
      @[simp]
      theorem Nat.choose_zero_right (n : ) :
      n.choose 0 = 1
      @[simp]
      theorem Nat.choose_zero_succ (k : ) :
      choose 0 k.succ = 0
      theorem Nat.choose_succ_succ' (n k : ) :
      (n + 1).choose (k + 1) = n.choose k + n.choose (k + 1)
      theorem Nat.choose_succ_left (n k : ) (hk : 0 < k) :
      (n + 1).choose k = n.choose (k - 1) + n.choose k
      theorem Nat.choose_succ_right (n k : ) (hn : 0 < n) :
      n.choose (k + 1) = (n - 1).choose k + (n - 1).choose (k + 1)
      theorem Nat.choose_eq_choose_pred_add {n k : } (hn : 0 < n) (hk : 0 < k) :
      n.choose k = (n - 1).choose (k - 1) + (n - 1).choose k
      theorem Nat.choose_eq_zero_of_lt {n k : } :
      n < kn.choose k = 0
      @[simp]
      theorem Nat.choose_self (n : ) :
      n.choose n = 1
      @[simp]
      theorem Nat.choose_succ_self (n : ) :
      n.choose n.succ = 0
      @[simp]
      theorem Nat.choose_one_right (n : ) :
      n.choose 1 = n
      theorem Nat.triangle_succ (n : ) :
      (n + 1) * (n + 1 - 1) / 2 = n * (n - 1) / 2 + n
      theorem Nat.choose_two_right (n : ) :
      n.choose 2 = n * (n - 1) / 2

      choose n 2 is the n-th triangle number.

      theorem Nat.choose_pos {n k : } :
      k n0 < n.choose k
      theorem Nat.choose_eq_zero_iff {n k : } :
      n.choose k = 0 n < k
      theorem Nat.choose_mul {n k s : } (hkn : k n) (hsk : s k) :
      n.choose k * k.choose s = n.choose s * (n - s).choose (k - s)
      theorem Nat.add_choose (i j : ) :
      (i + j).choose j = (i + j).factorial / (i.factorial * j.factorial)
      @[simp]
      theorem Nat.choose_symm {n k : } (hk : k n) :
      n.choose (n - k) = n.choose k
      theorem Nat.choose_symm_of_eq_add {n a b : } (h : n = a + b) :
      n.choose a = n.choose b
      theorem Nat.choose_symm_add {a b : } :
      (a + b).choose a = (a + b).choose b
      theorem Nat.choose_symm_half (m : ) :
      (2 * m + 1).choose (m + 1) = (2 * m + 1).choose m
      theorem Nat.choose_succ_right_eq (n k : ) :
      n.choose (k + 1) * (k + 1) = n.choose k * (n - k)
      @[simp]
      theorem Nat.choose_succ_self_right (n : ) :
      (n + 1).choose n = n + 1
      theorem Nat.choose_mul_succ_eq (n k : ) :
      n.choose k * (n + 1) = (n + 1).choose k * (n + 1 - k)
      def Nat.fast_choose (n k : ) :

      A faster implementation of choose, to be used during bytecode evaluation and in compiled code.

      Equations
        Instances For

          Inequalities #

          theorem Nat.choose_le_succ_of_lt_half_left {r n : } (h : r < n / 2) :
          n.choose r n.choose (r + 1)

          Show that Nat.choose is increasing for small values of the right argument.

          theorem Nat.choose_le_middle (r n : ) :
          n.choose r n.choose (n / 2)

          choose n r is maximised when r is n/2.

          Inequalities about increasing the first argument #

          theorem Nat.choose_le_add (a b c : ) :
          a.choose c (a + b).choose c
          theorem Nat.choose_le_choose {a b : } (c : ) (h : a b) :
          a.choose c b.choose c
          theorem Nat.choose_mono (b : ) :
          Monotone fun (a : ) => a.choose b

          Multichoose #

          Whereas choose n k is the number of subsets of cardinality k from a type of cardinality n, multichoose n k is the number of multisets of cardinality k from a type of cardinality n.

          Alternatively, whereas choose n k counts the number of combinations, i.e. ways to select k items (up to permutation) from n items without replacement, multichoose n k counts the number of multicombinations, i.e. ways to select k items (up to permutation) from n items with replacement.

          Note that multichoose is not the multinomial coefficient, although it can be computed in terms of multinomial coefficients. For details see https://mathworld.wolfram.com/Multichoose.html

          TODO: Prove that choose (-n) k = (-1)^k * multichoose n k, where choose is the generalized binomial coefficient. https://github.com/leanprover-community/mathlib/pull/15072#issuecomment-1171415738

          @[irreducible]
          def Nat.multichoose :

          multichoose n k is the number of multisets of cardinality k from a type of cardinality n.

          Equations
            Instances For
              @[simp]
              @[simp]
              theorem Nat.multichoose_zero_succ (k : ) :
              multichoose 0 (k + 1) = 0
              theorem Nat.multichoose_succ_succ (n k : ) :
              (n + 1).multichoose (k + 1) = n.multichoose (k + 1) + (n + 1).multichoose k
              @[simp]
              theorem Nat.multichoose_one (k : ) :
              @[simp]
              theorem Nat.multichoose_two (k : ) :
              multichoose 2 k = k + 1
              @[simp]
              @[irreducible]
              theorem Nat.multichoose_eq (n k : ) :
              n.multichoose k = (n + k - 1).choose k