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C programs QnA

Q. If sum of two integer is 14 and if one of them is 8 then find other integer. Ans: Well, this is very simple. Simply, one integer is 8 and sum of two integers is 14. So, we need to subtract the number 8 from their sum of 14 to get the second number. So, the second number is: 14 - 8 = 6 C-Program of this task Code---> #include<stdio.h> int main () { int num1 = 8 , num2 , sum = 14 ; num2 = sum - num1 ; printf ( "The second number is: %d" , num2 ); return 0 ; }
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nPr & nCr Calculator using C-Language



In this C program we will calculate the nPr or nCr and print the result, in the screen. The choice will be taken from the user. User can calculate nCr or nPr by choice.
If you are new in nPr or nCr and want to know about nPr or nCr,please, click below :
https://gradestack.com/Complete-CAT-Prep/Permutation-and/Meaning-and-Derivation-of/19142-3882-35788-study-wtw

input:

User's Choice. (i.e. 1.for nPr and 2.for nCr) & then, objects(n) and samples(r).

output:

The Calculated result will be printed on the screen.

CODE---->


#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<stdlib.h>
 int npr();
int ncr();
int fact(int num3);
main() {
     int choice;
     printf("*********Menu*********\n");
     printf("1.nPr\n2.nCr\n\n");
     printf("Please, Enter Your Choice : ");
     scanf("%d",&choice);
     if(choice==1)
     npr();
     if(choice==2)
     ncr();
     getch();
}
     int npr()
     {
         int num1,num2,i,nr,answer_1,answer_2,answer_3;
         printf("\nP(n,r) \n");
         printf("Enter n(objects) :" );
         scanf("%d",&num1);
         printf("\nEnter r(sample) : ");
         scanf("%d",&num2);
         answer_1=fact(num1);
         nr=num1-num2;
         answer_2=fact(nr);
         answer_3=answer_1/answer_2;
         printf("\n P(n,r) : %d",answer_3);
     }
     int ncr()
     {
   int num1,num2,i,nr,answer_1,answer_2,answer_3,answer_4,answer_5;
         printf("\nC(n,r) \n");
         printf("Enter n(objects) :" );
         scanf("%d",&num1);
         printf("\nEnter r(sample) : ");
         scanf("%d",&num2);
         answer_1=fact(num1);
         nr=num1-num2;
         answer_2=fact(nr);
         answer_3=fact(num2);
         answer_4=answer_2*answer_3;
         answer_5=answer_1/answer_4;
         printf("\n C(n,r) : %d",answer_5);
     }
     int fact(int num3)
     {
         int i,ans=1;
         for(i=1;i<=num3;i++)
         {
             ans=ans*i;
         }
         return ans;
         }



Don't just read, write it, run it.....

RESULT:


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