Question

Find the number of permutations x1, x2, x3, x4, x5 of numbers 1, 2, 3, 4, 5 such that the sum of five products x1x2x3 + x2x3x4 + x3x4x5 + x4x5x1 + x5x1x2 is divisible by 3

Answer 1

Answer:

좋은아침이에요저를 따라 오십시오.

jeoleul ttala osibsio.

안녕하세요

Answer 2

Answer:

C program for possible permutations.

Question :Find the number of permutations x1, x2, x3, x4, x5 of numbers 1, 2, 3, 4, 5 such that the sum of five products x1x2x3 + x2x3x4 + x3x4x5 + x4x5x1 + x5x1x2 is divisible by 3

Step-by-step explanation:

From the above question,

They have given :

The number of permutations x1, x2, x3, x4, x5 of numbers 1, 2, 3, 4, 5 such that the sum of five products x1x2x3 + x2x3x4 + x3x4x5 + x4x5x1 + x5x1x2 is divisible by 3.

Here is the program to find the permudation :

C PROGRAM :

#include <stdio.h>

void swap(char *x, char *y)

{

   char temp;

   temp = *x;

   *x = *y;

   *y = temp;

}

void permute(char *a, int l, int r)

{

  int i;

  if (l == r)

    printf("%s\n", a);

  else

  {

      for (i = l; i <= r; i++)

      {

         swap((a+l), (a+i));

         permute(a, l+1, r);

         swap((a+l), (a+i));

      }

  }

}

int main()

{

  char str[] = "ABC";

  int n = strlen(str);

  permute(str, 0, n-1);

  return 0;

}

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