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momodishev

Calculates n! / (k!*(n-k)!) where (1 < k < n < 100)

Apr 5th, 2014
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C# 1.20 KB | None | 0 0
  1. Console.WriteLine("This program calculates n! / (k!*(n-k)!) where (1 < k < n < 100).");
  2.         Console.Write("Please, enter n = : ");
  3.         int number = int.Parse(Console.ReadLine());
  4.         Console.Write("Please, enter k = : ");
  5.         int coeficient = int.Parse(Console.ReadLine());
  6.  
  7.         double multidevider = 1;
  8.         double subtractor = 1;
  9.         int subCounter = number - coeficient;
  10.         double faktoriel = 1;
  11.  
  12.         if (coeficient < number && coeficient > 1 && number < 100)
  13.         {
  14.             for (double count = 1; count <= number; count++)
  15.             {
  16.                 if (coeficient >= count)
  17.                 {
  18.                    
  19.                     multidevider *= count;
  20.                 }
  21.                 if (subCounter >= count)
  22.                 {
  23.                     subtractor *= count;
  24.                 }
  25.                 faktoriel = faktoriel * count;
  26.             }
  27.             double division = faktoriel / (multidevider * subtractor);
  28.             Console.WriteLine("The result of {1}!/({2}!*({1}-{2})!) is {0:0.00}", division, number, coeficient);
  29.  
  30.         }
  31.         else
  32.         {
  33.             Console.WriteLine("Invalid entry. (1 < k < n < 100)");
  34.         }
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