output_description: Print two lines.  
  The first line should contain the maximum possible arithmetic mean of the
  values of selected items. The output should be considered correct if the
  absolute or relative error is at most 10^{-6}.  
  The second line should contain the number of ways to select items so that the
  mean of the values of selected items is maximized.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      N A B
      v_1
      v_2
      ...
      v_N
attempt: import sys
  import math
  import collections
  import itertools
  import array
  import inspect
  from scipy.special import comb
  
  # Set max recursion limit
  sys.setrecursionlimit(1000000)
  
  
  # Debug output
  def chkprint(*args):
      names = {id(v): k for k, v in inspect.currentframe().f_back.f_locals.items()}
      print(", ".join(names.get(id(arg), "???") + " = " + repr(arg) for arg in args))
  
  
  # Binary converter
  def to_bin(x):
      return bin(x)[2:]
  
  
  def li_input():
      return [int(_) for _ in input().split()]
  
  
  def gcd(n, m):
      if n % m == 0:
          return m
      else:
          return gcd(m, n % m)
  
  
  def gcd_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = gcd(v, L[i])
  
      return v
  
  
  def lcm(n, m):
      return (n * m) // gcd(n, m)
  
  
  def lcm_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = lcm(v, L[i])
  
      return v
  
  
  # Width First Search (+ Distance)
  def wfs_d(D, N, K):
      """
      D: 隣接行列(距離付き)
      N: ノード数
      K: 始点ノード
      """
  
      dfk = [-1] * (N + 1)
      dfk[K] = 0
  
      cps = [(K, 0)]
      r = [False] * (N + 1)
      r[K] = True
      while len(cps) != 0:
          n_cps = []
          for cp, cd in cps:
              for i, dfcp in enumerate(D[cp]):
                  if dfcp != -1 and not r[i]:
                      dfk[i] = cd + dfcp
                      n_cps.append((i, cd + dfcp))
                      r[i] = True
  
          cps = n_cps[:]
  
      return dfk
  
  
  # Depth First Search (+Distance)
  def dfs_d(v, pre, dist):
      """
      v:  現在のノード
      pre: １つ前のノード
      dist: 現在の距離
  
      以下は別途用意する
      D: 隣接リスト(行列ではない)
      D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
      """
  
      global D
      global D_dfs_d
  
      D_dfs_d[v] = dist
  
      for next_v, d in D[v]:
          if next_v != pre:
              dfs_d(next_v, v, dist + d)
  
      return
  
  
  def sigma(N):
      ans = 0
      for i in range(1, N + 1):
          ans += i
      return ans
  
  
  class Combination:
      def __init__(self, n, mod=10**100):
  
          g1 = [1, 1]
          g2 = [1, 1]
          inverse = [0, 1]
          for i in range(2, n + 1):
              g1.append((g1[-1] * i) % mod)
              inverse.append((-inverse[mod % i] * (mod // i)) % mod)
              g2.append((g2[-1] * inverse[-1]) % mod)
          self.MOD = mod
          self.N = n
          self.g1 = g1
          self.g2 = g2
          self.inverse = inverse
  
      def __call__(self, n, r):
          if r < 0 or r > n:
              return 0
          r = min(r, n - r)
          return self.g1[n] * self.g2[r] * self.g2[n - r] % self.MOD
  
  
  # --------------------------------------------
  
  dp = None
  
  
  def main():
      N, A, B = li_input()
      V = li_input()
  
      V.sort(reverse=True)
  
      maxavg = sum(V[:A]) / A
      V_ = V[:A]
  
      if V_[0] == V_[-1]:
          n = V.count(V_[0])
  
          ans = 0
          for r in range(A, min(n + 1, B + 1)):
              ans += comb(n, r)
  
          print(maxavg)
          print(int(ans))
  
      else:
          n = V.count(V_[-1])
          r = V_.count(V_[-1])
  
          ans = comb(n, r)
  
          print(maxavg)
          print(int(ans))
  
  
  main()
  
problem_description: Statement
  
  You are given N items.  
  The _value_ of the i-th item (1 \leq i \leq N) is v_i.  
  Your have to select at least A and at most B of these items.  
  Under this condition, find the maximum possible arithmetic mean of the values
  of selected items.  
  Additionally, find the number of ways to select items so that the mean of the
  values of selected items is maximized.
samples: [{"input": "5 2 2\n    1 2 3 4 5", "output": "4.500000\n    1\n    \n\nThe mean of the values of selected items will be maximized when selecting the\nfourth and fifth items. Hence, the first line of the output should contain\n4.5.  \nThere is no other way to select items so that the mean of the values will be\n4.5, and thus the second line of the output should contain 1.\n\n* * *"}, {"input": "4 2 3\n    10 20 10 10", "output": "15.000000\n    3\n    \n\nThere can be multiple ways to select items so that the mean of the values will\nbe maximized.\n\n* * *"}, {"input": "5 1 5\n    1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996", "output": "1000000000000000.000000\n    1"}]