Write two functions called permutation() and combination(), where permutation() computes and returns n!/(n-r)!whereas combination() returns. n!/(n-r)!r!A third function called par_input() gets the input values for n and r. An example of correct program behavior follows: Input: 3 2 Output: The permutation is 6, and the combination is 3.

Here is the program in C.

#include<stdio.h>   // header file used to input output operations

int permutation(int n, int r);  // function to computer permutation

int combination(int n, int r);  //function to compute combination

int factorial(int number);  // function to computer factorial

void par_input(int n, int r);    // function take take input value of n and r

int main()  //start of main() function body

{    

   int n,r;  // declare n and r variables

   par_input(n,r);  //calls par_input() function to get values of n and r

}

int permutation(int n, int r)   // method to calculate permutation

{

   return factorial(n) / factorial(n-r);  //formula for permutation

}

int combination(int n, int r)  // method to calculate combination

{

   return permutation(n, r) / factorial(r); //formula for combination

}

int factorial(int number)  // to find factorial of a number

{

   int fact = 1;      

   while(number > 0)  //loop continues until value of number gets <=0

   {

       fact = fact *number;  

       number--;

   }      

   return fact;

}

void par_input(int n, int r)  //function to take input values

{

   printf("Enter n: ");  //prompts user to enter the value of n

   scanf("%d", &n);      //reads the value of n from user

   printf("Enter r: "); ////prompts user to enter the value of r

   scanf("%d", &r); //reads the value of r from user

  //calls the combination and permutation methods to display results

   printf("The permutation is = %d\n", permutation(n, r));    

   printf("The Combination is = %d", combination(n, r));      

}

Explanation:

The output of the program is attached in the screen shot.

i think the answer is b. sql.

hope it