C Exercises: Number of combinations of a, b, c and d (0 <= a, b, c, d <= 9) where (a + b + c + d) will be equal to n

Last update on May 27 2022 12:13:29 (UTC/GMT +8 hours)

C Basic Declarations and Expressions: Exercise-135 with Solution

Write a C program which reads an integer n and find the number of combinations of a, b, c and d (0 <= a, b, c, d <= 9) where (a + b + c + d) will be equal to n.

Input:
n (1 <= n <= 50)

Sample Solution:

C Code:

#include <stdio.h>
int main() {
  int i, j, k, l, n;
  printf("Input a number:\n");
  scanf("%d", & n);
  if (n >= 1 && n <= 39)
    printf("\na + b + c + d = n"); {
    int count = 0;
    for (i = 0; i <= 9; i++) {
      for (j = 0; j <= 9; j++) {
        for (k = 0; k <= 9; k++) {
          for (l = 0; l <= 9; l++)

          {
            if (i + j + k + l == n) {
              printf("\n%d, %d, %d, %d", i, j, k, l);
              count++;
            }
          }

        }

      }
    }
    printf("\n\nTotal number of combinations:");
    printf("\n%d\n", count);
  }
  return 0;
}

Sample Output:

Input a number:
5

a + b + c + d = n
0, 0, 0, 5
0, 0, 1, 4
0, 0, 2, 3
0, 0, 3, 2
0, 0, 4, 1
0, 0, 5, 0
0, 1, 0, 4
0, 1, 1, 3
0, 1, 2, 2
0, 1, 3, 1
0, 1, 4, 0
0, 2, 0, 3
0, 2, 1, 2
0, 2, 2, 1
0, 2, 3, 0
0, 3, 0, 2
0, 3, 1, 1
0, 3, 2, 0
0, 4, 0, 1
0, 4, 1, 0
0, 5, 0, 0
1, 0, 0, 4
1, 0, 1, 3
1, 0, 2, 2
1, 0, 3, 1
1, 0, 4, 0
1, 1, 0, 3
1, 1, 1, 2
1, 1, 2, 1
1, 1, 3, 0
1, 2, 0, 2
1, 2, 1, 1
1, 2, 2, 0
1, 3, 0, 1
1, 3, 1, 0
1, 4, 0, 0
2, 0, 0, 3
2, 0, 1, 2
2, 0, 2, 1
2, 0, 3, 0
2, 1, 0, 2
2, 1, 1, 1
2, 1, 2, 0
2, 2, 0, 1
2, 2, 1, 0
2, 3, 0, 0
3, 0, 0, 2
3, 0, 1, 1
3, 0, 2, 0
3, 1, 0, 1
3, 1, 1, 0
3, 2, 0, 0
4, 0, 0, 1
4, 0, 1, 0
4, 1, 0, 0
5, 0, 0, 0

Total number of combinations:
56

Flowchart:

C programming Code Editor:

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C Programming: Tips of the Day

Static variable inside of a function in C

The scope of variable is where the variable name can be seen. Here, x is visible only inside function foo().

The lifetime of a variable is the period over which it exists. If x were defined without the keyword static, the lifetime would be from the entry into foo() to the return from foo(); so it would be re-initialized to 5 on every call.

The keyword static acts to extend the lifetime of a variable to the lifetime of the programme; e.g. initialization occurs once and once only and then the variable retains its value - whatever it has come to be - over all future calls to foo().

Ref : https://bit.ly/3fOq7XP