Combination formula "n! / ((n-k)!*k!)" doesn't work [closed] - c

Question

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.
Closed 10 years ago.
why is = n! / ((n-k)!*k!) not printing?
Also will this code solve the problem below?
stuck.
"The number of combinations of n things taken k at a time as an integer"
A little more clarification: "For example, the combinations of four items a,b,c,d taken three at a time are abc, abd, acd, and bcd. In other words, there are a total of four different combinations of four things "taken three at a time"."
#include <stdio.h>
#include <math.h>
int main (void)
{
int z = 0, n_in, k_in, k = 0, n = 0, result, nfr = 0, kfr = 0;
do
{
printf("Enter the number of items in the list (n):");
scanf("%d*c", &n_in);
if (n_in>1 && n_in<11)
{
printf("Enter the number of items to choose (k)");
scanf("%d*c", &k_in);
if (k_in>0 && k_in<5)
{
if (k_in <= n_in)
{
k_in = k;
n_in = n;
result = n! / ((n-k)!*k!);
z = 1;
}
else
printf("?Please Try again k must be less than n \n");
}
else
printf("?Invalid input: Number must be between 1 and 4 \n");
}
else
printf("?Invalid input: Number must be between 1 and 10 \n");
} while (z == 0);
result = (nfr / (nfr * kfr));
printf("k value = %d n value = %d the result is %d", nfr, kfr, result);
return 0;
}

Answer 1

This line:
result = n! / ((n-k)!*k!);
...is not valid C code. ! in C means "not".
You will need to provide a factorial function so that you can call:
result = factorial(n) / (factorial(n-k) * factorial(k));

Answer 2

! is not the NOT operator in C. Use this factorial function instead.
int factorial(int n)
{
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
So your calculation would be:
result = factorial(n) / (factorial(n-k)*factorial(k));
There are probably faster ways to do it, but this is readable.
Also, this line
result = (nfr / (nfr * kfr));
Does not make any sense to me, since both nfr and kfr are zero, but I guess you wanted to get the code to compile, before completing the logic.
EDIT:
complete code should look like this:
#include <stdio.h>
#include <math.h>
int factorial(int n)
{
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
int main (void)
{
int z = 0, n_in, k_in, k = 0, n = 0, result, nfr = 0, kfr = 0;
do
{
printf("Enter the number of items in the list (n):");
scanf("%d*c", &n_in);
if (n_in>1 && n_in<11)
{
printf("Enter the number of items to choose (k)");
scanf("%d*c", &k_in);
if (k_in>0 && k_in<5)
{
if (k_in <= n_in)
{
k_in = k;
n_in = n;
result = factorial(n) / (factorial(n-k)*factorial(k));
//result = n! / ((n-k)!*k!);
z = 1;
}
else
printf("?Please Try again k must be less than n \n");
}
else
printf("?Invalid input: Number must be between 1 and 4 \n");
}
else
printf("?Invalid input: Number must be between 1 and 10 \n");
} while (z == 0);
//result = (nfr / (nfr * kfr));
printf("k value = %d n value = %d the result is %d\n", nfr, kfr, result);
return 0;
}
Output:
~/so$ gcc test.cc
~/so$ ./a.out
Enter the number of items in the list (n):3
Enter the number of items to choose (k)2
k value = 0 n value = 0 the result is 1
~/so$