Combinatoric selections

Author: adamadair

problem_053/problem53.md

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+# Combinatoric selections
+# Problem 53
+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?
+
+# Solution Notes
+I take advantage of the large number capabilities of Ruby in my solution. The overview
+from Project Euler provides some very complex mathematics that could also provide the 
+soluton. I might analyze this and perhaps code it in a differenct language later.

problem_053/solution.rb

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+class Combinatorics
+  def initialize 
+    @factorial = []
+    @factorial<<1 # 0 factorial is 1 
+    @factorial<<1
+    (2..100).to_a.each{|n|@factorial<<(1..n).to_a.reduce(:*)}
+  end
+  def calc(n,r)
+    @factorial[n]/(@factorial[r]*@factorial[n-r])  
+  end
+end
+
+c = Combinatorics.new
+count = 0
+(1..100).to_a.each{|n|
+  (1..100).to_a.each{|r|
+    count += 1 if c.calc(n,r)>1_000_000
+  }
+}
+puts count