Question

In Probability, number of"combinations" (sometimes referred to as binomial coefficients) ofn times taken r at a time is written as C (n.r) = | | = . If an order conscious subset of r times taken from a set of n times- the permutations, it is written as P(n, r) = . Both cases, r n 1. n! r r(nr)! n! Create a script that will prompt the user to type in values for n and r , then calculate C (n,r) and P(n,). After the display of the result, the user should be prompted as well whether shehe wants to repeat the process or not. Requirements 1) The user should be prompted 2 times for n and r 2) Check whether r S n. If this condition is not satisfied, warn the user to type in 3) Create a user defined function that computes the Clo.r). (This user defined function 4) 5) Store result of Cln.land Pín) in an array. appropriate values for r and n, AGAIN should take 2 arquments and return the numerical result of the Clo.r) Create a user defined function that computes the Pin ) (This user defined function should take 2 arquments and return the numerical result of the P(n,r).) After the display of the result, the user should be prompted as well whether she he wants to repeat the process or not. Once the user typed in her/his decision of "NOT REPEATING THE PROCESS", the stored values in the array should be displayed all together For displaying the result array, you should use "dereferencing the pointer" approach. 6) 7) 8)

Use C programming,

Also please use “printf” function when displaying text

Answer 1

C Program:

#include <stdio.h>

  

long fact(int n) {

   int c;

   long result = 1;

   for (c = 1; c <= n; c++)

      result = result*c;

   return result;

}

  

long C(int n, int r) {

   long result;    

   result = fact(n)/(fact(r)*fact(n-r));

   return result;

}

  

long P(int n, int r) {

   long result;

   result = fact(n)/fact(n-r);

   return result;

}

  

  

int main(){

   int n[100], r[100], j, i=0;

   long ncr[100], npr[100];

   char ch;

   do{

    do{

        printf("Enter the value of n: ");

        scanf("%d",&n[i]);

        printf("Enter the value of r: ");

        scanf("%d",&r[i]);

        if(r[i]>n[i])

            printf("n and r should be r<=n try agian:\n");

    }while(r[i]>n[i]);

    ncr[i] = C(n[i], r[i]);

    npr[i] = P(n[i], r[i]);

    i++;

    printf("Do you want to continue (Y/N): ");

    scanf(" %c",&ch);

   }while(ch == 'y' || ch == 'Y');

   for(j=0; j<i; j++){

        printf("%dC%d = %ld\n", n[j], r[j], ncr[j]);

        printf("%dP%d = %ld\n", n[j], r[j], npr[j]);

   }

   return 0;

}

Output: