James242 0 Точки

## 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");

Console.WriteLine("Enter the second digit to muliply");
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);
}

}

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

Тагове:
0
Slavi007 0 Точки

Hello, James.

I will try to show the mathematic logic on the problem you have.

If we examine your code which is absolutely correct we can see that if k is 2 and n is 4 the code methodically calculates n!/k!

For example (using your code): we type n= 4 and k=2, then the code calculates n! = 24 and k! = 2 and then return n!/k! = 12 and that is correct.

The second solution is also correct and it is also quite faster to be compiled.

For example (using the second solution) we type n=4 and k=2 then the for loop starts and here comes the magic. i starts from k+1 which is 3 in this example and i is added to result = result*i; Then i = 4 and it is again added to result which is now 3 and now becomes 3*4 =12;

You can try with other numbers but the result will be always n!/k!