Re: puzzle with string permutations [photo]

[Emanuel Berg]

Yuri Khan wrote:

> In general, with a multiset {X_1:n_1, ..., X_k:n_k}, if we
> number each letter, we get (n_1 + ... + n_k)! permutations,
> and then we have to adjust for duplicates for each
> individual letter by dividing by (n_1)! ... (n_k)!.
>
> n_perm = (n_1 + ... + n_k)! / (n_1)! ... (n_k)!

You mean like this?

(require 'cl-lib)

(defun cl-faculty (n)
  (cl-loop with prod = 1
    for i from 2 to n do
    (setq prod (* i prod))
    finally return prod) )
;; (cl-faculty  5) ;       120
;; (cl-faculty 10) ; 3 628 800

(defun count (e l)
  (seq-count (lambda (elem) (= elem e)) l) )
;; (count ?o '(?d ?g ?o ?o)) ; 2

(defun product-string (str)
  (let*((str-list         (string-to-list str))
        (str-list-no-dups (cl-remove-duplicates str-list))
        (prod 1) )
    (dolist (e str-list-no-dups)
      (setq prod (* prod (cl-faculty (count e str-list)))) )
    prod ))
;; (product-string "ogod") ; 2

(defun perms-string-num (str)
  (let ((n (cl-faculty (length str)))
        (r (product-string str)) )
    (/ n r) ))
;; (perms-string-num "ogod")   ;  12
;; (perms-string-num "kudtce") ; 720

-- 
underground experts united
https://dataswamp.org/~incal