Matlab commands:

>> a=string(sortrows(uniqueperms('perplexus')));

>> find(a=="perplexus")

ans =

       32670

       

       

A more complex way--VB program:

Private Sub Form_Load()

  x$ = "EELPPRSUX": n = 1

  Do While x$ <> "PERPLEXUS"

    permute x$

    n = n + 1

  Loop

  Text1.Text = x$ & " " & n

End Sub

which invokes the routine that's the solution to my Permutations puzzle.

It finds:

PERPLEXUS 32670

If there had been no repeated letters in the word, the method described in my solution to Permutation List would have been directly applicable.  Modifying it a little to take care of the repeated letters:

Alphabetic order EELPPRSUX

Those that start with E or L all come before PERPLEXUS:

With E:  8!/2 = 20160

With L:  8!/4 = 10080

Those that start with PE followed next by E, L or P also precede PERPLEXUS:

PEE: 6! = 720

PEL: 6! = 720

PEP: 6! = 720

Those that start PER, but then continue with E or L:

PERE: 5! = 120

PERL: 5! = 120

Those that start PERP, but then continue with E:

PERPE: 4! = 24

Now, those that start PERPLE, have endings

SUX, USX, UXS, USX or XSU all before XUS.

That's 5 more, and then we add 1 because we want PERPLEXUS's ordinal number.

20160 + 10080 + 3*720 + 2*120 + 24 + 5 + 1 = 32670

I'll admit: the first time through this analytic version, I forgot to consider the sequences that start PEP, and wound up 720 short, which I wouldn't have noticed had I done the analytic version without the computer version.