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Consecutive Placement - Not (Posted on 2014-12-01) Difficulty: 3 of 5
Positive integers from 1 to 8 inclusively are randomly placed on the faces of a regular octahedron with each face being assigned a different number. For the purposes of the puzzle, 8 and 1 are considered as consecutive numbers.

Determine the probability that no two consecutive numbers are placed on faces that share an edge.

No Solution Yet Submitted by K Sengupta    
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Solution computer assisted solution Comment 1 of 1
Any octahedron with randomly placed digits 1 - 8 can be oriented so the 1 is on top, and we can consider the three faces adjacent to that one starting with the highest numbered and proceding left to right (counterclockwise seen from the top). Also, mirror images are not counted, as the probability is the same for those, by requiring the number to the right of the highest, immediately below the top face, to be higher than the one to the left. There are 840 equally likely number placements (7!/3!), 10 of which satisfy the criterion looked for in the probability.  Those 10 are shown here:

1
5 4 3
 7 6 8
2

1
5 4 3
 8 6 7
2

1
6 4 3
 2 7 8
5

1
6 5 3
 2 7 8
4

1
7 4 3
 2 6 5
8

1
7 5 4
 3 8 2
6

1
7 6 4
 3 8 2
5

1
7 6 5
 4 2 3
8

1
7 6 5
 4 3 2
8

1
7 6 3
 4 8 5
2

10 840 0.01190476190

The probability thus is 10/840 = 1/84 ~= 0.01190476190.

DefDbl A-Z
Dim crlf$, adj$(7), good, o$


Function mform$(x, t$)
  a$ = Format$(x, t$)
  If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$
  mform$ = a$
End Function

Private Sub Form_Load()
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True
 DoEvents
 
'    1
' 2  4  6
'   3  5  7
'      8
 adj$(1) = "246"
 adj$(2) = "37"
 adj$(3) = "48"
 adj$(4) = "5"
 adj$(5) = "68"
 adj$(6) = "7"
 adj$(7) = "8"
 
 
 remain$ = "2345678": h$ = remain$
 Do
  o$ = "1" + remain$
  If Mid(o$, 2, 1) > Mid(o$, 4, 1) And Mid(o$, 2, 1) > Mid(o$, 6, 1) Then
   If Mid(o$, 4, 1) > Mid(o$, 6, 1) Then
    good = 1
    For i = 1 To 7
      checkFace i
    Next
    If good Then
      goodct = goodct + 1
      Text1.Text = Text1.Text & Mid(o$, 1, 1) & crlf
      Text1.Text = Text1.Text & Mid(o$, 2, 1) & " " & Mid(o$, 4, 1) & " " & Mid(o$, 6, 1) & crlf
      Text1.Text = Text1.Text & " " & Mid(o$, 3, 1) & " " & Mid(o$, 5, 1) & " " & Mid(o$, 7, 1) & crlf
      Text1.Text = Text1.Text & Mid(o$, 8, 1) & crlf & crlf
    End If
    ct = ct + 1
   End If
  End If
  permute remain$
 Loop Until remain$ = h$
 Text1.Text = Text1.Text & goodct & Str(ct) & mform(goodct / ct, " 0.00000000000") & crlf

End Sub

Sub checkFace(wh)
  For i = 1 To Len(adj$(wh))
    wh2 = Val(Mid(adj$(wh), i, 1))
    v1 = Val(Mid(o$, wh, 1))
    v2 = Val(Mid(o$, wh2, 1))
    diff = Abs(v2 - v1)
    If diff = 1 Or diff = 7 Then good = 0: Exit For
  Next
End Sub


  Posted by Charlie on 2014-12-01 16:23:18
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