Abort trap 6 error in C

You are writing to memory you do not own:

int board[2][50]; //make an array with 3 columns  (wrong)
                  //(actually makes an array with only two 'columns')
...
for (i=0; i<num3+1; i++)
    board[2][i] = 'O';
          ^

Change this line:

int board[2][50]; //array with 2 columns (legal indices [0-1][0-49])
          ^

To:

int board[3][50]; //array with 3 columns (legal indices [0-2][0-49])
          ^

When creating an array, the value used to initialize: [3] indicates array size.
However, when accessing existing array elements, index values are zero based.

For an array created: int board[3][50];
Legal indices are board[0][0]...board[2][49]

EDIT To address bad output comment and initialization comment

add an additional "\n" for formatting output:

Change:

  ...
  for (k=0; k<50;k++) {
     printf("%d",board[j][k]);
  }
 }

       ...

To:

  ...
  for (k=0; k<50;k++) {
     printf("%d",board[j][k]);
  }
  printf("\n");//at the end of every row, print a new line
}
...  

Initialize board variable:

int board[3][50] = {0};//initialize all elements to zero

( array initialization discussion... )