C# Book chapter 6 exercise 6 - Dividing by 2 factorials
Hello! I do not speak Bulgarian, but I have finished reading the Fundamentals of Computer Programming with C# book, am going back over some of the exercises I did not initally complete, and am hoping this forum might be useful to me even though I don't speak Bulgarian. I know a lot of you must speak English since much of the course material is in English. I just noticed the language toggle at the bottom of the website's page.
Chapter 6 exercise 6 asks us to "write a program that calculates N!/K! for given N and K (1<K<N).
I wrote a solution to this problem with a calculateFactorial function that is used to calculate N! and then K!, after which those values are divided.
    static void Main(string[] args)
        {
            Console.WriteLine("Enter the first digit to muliply");
            BigInteger n = BigInteger.Parse(Console.ReadLine());
            Console.WriteLine("Enter the second digit to muliply");
            BigInteger k = BigInteger.Parse(Console.ReadLine());
            if (1 > k || k > n)
            {
                Console.WriteLine("second number must be larger than 1 and smaller than first number.");
            }
            else
            {
                BigInteger nFactorial = calculateFactorial(n);
                BigInteger kFactorial = calculateFactorial(k);
                Console.WriteLine("n! = " + nFactorial);
                Console.WriteLine("k! = " + kFactorial);
                BigInteger output = nFactorial / kFactorial;
                Console.WriteLine("{0} / {1} = {2}", nFactorial, kFactorial, output);
            }
           
            Console.ReadLine();
        }
        static BigInteger calculateFactorial(BigInteger x)
        {
            BigInteger factorial = 1;
            do
            {
                factorial *= x;
                x--;
            }
            while (x > 0);
            return factorial;
        }
    }
However, I'm looking at the word .doc with the solutions and this provided solution is more elegant since it only caluclates k+1 * each n and removes the division step.
 for (int i = k+1; i <= n; i++)
                {
                    result *= i;
                }
I don't quite understand the mathematical logic on why doing the above caluclation is the same as N!/K!
Any insight on this topic would be very much appreciated!
James
Love it! Thanks so much for explaining this to me in English! Nice people in this forum.