fill any 6 of 18 cells randomly with an "X"

I need a macro that will do exactly as the title says. I have 18 cells that, when a command button is pressed, it will randomly pick 6 of the 18 cells and fill them with an x. Is this even possible?

can use the below

 Sub Randmnumber()

Range("A1:Z200").ClearContents

For TM = 1 To 6

RN = Int((6 - 18 + 1) * Rnd + 18)
CN = Int((6 - 18 + 1) * Rnd + 18)

If RN <> 0 And CN <> 0 Then
Cells(RN, CN).Value = "x"
Else
TM = TM - 1
End If


Next


End Sub

Another way:

Sub fill_random_x()
Dim x As Integer, y As Integer
Dim rng As Range
Set rng = Sheets("Sheet1").Range("A1:C6")
    
rng.ClearContents
   
Do While Application.WorksheetFunction.CountIf(rng, "x") < 6
    x = Application.WorksheetFunction.RandBetween(1, rng.Rows.Count)
    y = Application.WorksheetFunction.RandBetween(1, rng.Columns.Count)
    Cells(x, y) = "x"
Loop
    
   
End Sub

Sub fill_random_x_v2()
Dim x As Integer, y As Integer
Dim rng As Range
Set rng = Sheets("Sheet1").Range("A1:C6")
   
rng.ClearContents
   
Do While Application.WorksheetFunction.CountIf(rng, "x") < 6
    x = Int(1 + Rnd() * rng.Rows.Count)
    y = Int(1 + Rnd() * rng.Columns.Count)
    Cells(x, y) = "x"
Loop
   
End Sub

Above solution is not good.
It return wrong result on large number of results.

Here is distribution of previous solution:

4,60%	5,63%	6,42%
7,03%	5,70%	4,05%
6,64%	4,45%	5,55%
4,02%	6,96%	5,65%
5,53%	6,71%	4,50%
5,65%	4,03%	6,88%

And this is new solution:

5,53%	5,58%	5,59%
5,53%	5,59%	5,53%
5,56%	5,57%	5,54%
5,51%	5,55%	5,53%
5,57%	5,57%	5,58%
5,55%	5,58%	5,54%

Looks like RANDEBETWEEN function doesn0t work fine

Code is below