M is the number of six-element subsets that can be chosen from the set of the first 15 positive integers so that at least three of the six numbers are consecutive.
Find M.
Method 1:
The program counts all possible locations on the 1-15 integral number line for the triplet and three separately chosen numbers for the other three.
To prevent two separate triplets or any quadruplets, quintuplets or sextuplets from being counted more than once, the triplet that's chosen as the specific triplet must be the first triplet encountered. That is, for example 2,3,4,7,8,9 is counted only when the 2,3,4 is the dedicated triplet and 7, 8 and 9 are the individual numbers chosen, not the other way around. The individual numbers themselves are in order a < b < c, so permutations of those are not counted separately either.
The result is 2155 by method 1.
But method 2 counts only 550:
Method 2:
Just look at all the subsets and count how many meet the criterion.
It finds only 550.
Somehow the first method must be finding some subsets more than once, despite the precautions taken. Hmmm...
DefDbl A-Z
Dim crlf$
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr(13) & Chr(10)
For triple = 1 To 13
For a = 1 To 13
If a < triple Or a > triple + 2 Then
For b = a + 1 To 14
If b < triple Or b > triple + 2 Then
For c = b + 1 To 15
If c < triple Or c > triple + 2 Then
s$ = Space$(15)
Mid(s, triple, 3) = "xxx"
Mid(s, a, 1) = "x"
Mid(s, b, 1) = "x"
Mid(s, c, 1) = "x"
ix = InStr(s, "xxx")
If ix = triple Then
ct = ct + 1
End If
End If
Next
End If
Next
End If
Next
Next
Text1.Text = Text1.Text & crlf & ct & " done method 1"
ct = 0
For a = 1 To 10
For b = a + 1 To 11
For c = b + 1 To 12
For d = c + 1 To 13
For e = d + 1 To 14
For f = e + 1 To 15
hit = 0
If b = a + 1 Then cons = 1 Else cons = 0
If c = b + 1 Then
cons = cons + 1
If cons = 3 Then hit = 1
Else
cons = 0
End If
If d = c + 1 Then
cons = cons + 1
If cons = 3 Then hit = 1
Else
cons = 0
End If
If e = d + 1 Then
cons = cons + 1
If cons = 3 Then hit = 1
Else
cons = 0
End If
If f = e + 1 Then
cons = cons + 1
If cons = 3 Then hit = 1
Else
cons = 0
End If
If hit Then ct = ct + 1
Next
Next
Next
Next
Next
Next
Text1.Text = Text1.Text & crlf & ct & " done method 2"
End Sub
2155 done method 1
550 done method 2
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Posted by Charlie
on 2016-02-09 16:06:19 |
2 computer solutions; the second is probably right
