# Problem 53: Combinatoric selections
# There are exactly ten ways of selecting three from five, 12345:
# 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
# In combinatorics, we use the notation, 5C3 = 10.
# In general,
# nCr = n!/r!(n−r)!,where r ≤ n, n! = n×(n−1)×...×3×2×1, and 0! = 1.
# It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
# How many, not necessarily distinct, values of nCr, for 1 ≤ n ≤ 100, are greater than one-million?
from math import factorial
def nchoosek(n,k):
  return int(factorial(n))/(int(factorial(k))*int(factorial(n-k)))
S=0
for n in range(23,101):
  for k in range(1,n+1):
    if nchoosek(n,k)>1000000:
      S+=1
print(S)
  
