Question

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

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:

Enter the value of n: 2 Enter the value of r: 4 n and r should be r<=n try agian: Enter the value of n: 4 Enter the value of r: 2 Do you want to continue (Y/N): Y Enter the value of n: 4 Enter the value of r: 3 Do you want to continue (Y/N): n 4C2 = 6 4P2 = 12 403 = 4 4P3 = 24