nminusr = n-r;
for (i = 1; i<=n; i++)
{
a = a*i;
}
printf("a: %d\n", a);
for (j = 1; j<=r; j++)
{
b = b*j;
}
printf("b: %d\n", b);
for (k = 1; k<=nminusr; k++)
{
c = c*k;
}
printf("c: %d\n", c);
ncr = a/(b*c);
// Input: 6 3
// Output: 20
The given code calculates the value of nCr, which represents the number of combinations of n items taken r at a time (i.e., the number of ways to choose r items from a set of n items without considering the order). The code takes two integer inputs: 'n' and 'r' and calculates the value of nCr using the formula n! / (r! * (n-r)!). The code description is as follows:
1. Initialize variables: `nminusr` to store the value of (n-r), `a`, `b`, and `c` to store the factorial of 'n', 'r', and 'n-r', respectively.
2. Calculate the factorial of 'n' using a loop that runs from 1 to 'n' and multiplies the value of 'a' by each number in the loop.
3. Calculate the factorial of 'r' using a loop that runs from 1 to 'r' and multiplies the value of 'b' by each number in the loop.
4. Calculate the factorial of 'n-r' using a loop that runs from 1 to 'nminusr' and multiplies the value of 'c' by each number in the loop.
5. Calculate the value of nCr using the formula: nCr = a / (b * c).
6. Print the value of 'a', 'b', and 'c'.
7. Print the final result of nCr.
Example input: n = 6, r = 3
Output: nCr = 20
Note: It is assumed that the variables `a`, `b`, `c`, `nminusr`, `i`, `j`, and `k` have been properly declared and initialized before this code segment. The code does not handle input validation or edge cases, so it is assumed that the input values are positive integers where r <= n.
