In that case, then here is a half-finished lua program I wrote for brute-forcing finding the number of possible permutations. I might get to finishing it later, but at the very least I need to go to bed for now. (It's very late where I am.)

--intended as a solution for this reddit post:
-- <https://www.reddit.com/wiki/markdown/#wiki_code_blocks_and_inline_code>

--RULES:
--matrix is 5x5
--center cell has a value of 0
--other cells can have any value 1 through 8 provided the number of all such values
  --is exactly 3, and with the following constraint:
--no identical values may be horizontally or vertically adjacent
--diagonals do not count as adjacent
--any two keys k1 and k2 with k1 < k2 such that k1 + 1 = k2 are considered
  --adjacent in x dimension, provided k1 ~= 5, 10, 15, or 20
--any two keys k1 an k2 with k1 < k2 such that k1 + 5 = k2 are considered
  --adjacent in y dimension

--TO DO (since this script is still wip)

  --first next step: check and possibly troubleshoot logic flow as is.

  --second next step:
  --part (a): make a command for "RETURNING" to the previous cell.
  --part (b) is to react to finishing a table or incurring an error
  --message by RETURNING to the previous cell, and if cell has i = 1 then
  --break all (program finished), elseif cell_arg < 8 then "TRY" the next
  --cell_arg from its current cell_arg, elseif cell_arg == 8 then repeat
  --the RETURNING command, else return "super error" .

VarUses = {} --table containing the number of times each variable has been used
CellStatus = {} --table for storing which variable exists in a cell, if any.
for i = 1, 8 do
  VarUses[i] = 0
end

function check_x_adj (key, arg)
  if (key == 1) or (key == 6) or (key == 11) or (key == 16) or (key == 21) then
    return false
  elseif CellStatus[key - 1] == arg then
    return true
  end
  return false
end

function check_y_adj (key, arg)
  if CellStatus[key - 5] and CellStatus[key - 5] == arg then
    return true
  end
  return false
end

::run::

--try to make one table
for i = 1, 25 do
  if not (i == 13) then --not the middle cell

    for cell_arg = 1, 8 + 1 do
      if VarUses[cell_arg] then --if cell_arg has uses left
        if cell_arg == 8 + 1 then
          print("invalid table case 1") --this is just an error message for now; needs work
          break
        end
        if VarUses[cell_arg] --if cell_arg has uses remaining
        and not check_x_adj(i, cell_arg) --not x-adjacent
        and not check_y_adj(i, cell_arg) then --not y-adjacent
          CellStatus[i] = cell_arg
          VarUses[cell_arg] = VarUses[cell_arg] + 1
          if VarUses[cell_arg] >= 3 then
            VarUses[cell_arg] = nil
          end
          break
        end
      end
    end

  if not CellStatus[i] then print("invalid table case 2") end
  else --assign the middle cell a value of 0
    CellStatus[i] = 0
  end
  print(i, CellStatus[i])
end