Exercise 4.51: Implement a new kind of
assignment called permanent-set! that is not undone upon failure.  For
example, we can choose two distinct elements from a list and count the number
of trials required to make a successful choice as follows:


(define count 0)
(let ((x (an-element-of '(a b c)))
      (y (an-element-of '(a b c))))
  (permanent-set! count (+ count 1))
  (require (not (eq? x y)))
  (list x y count))

;;; Starting a new problem
;;; Amb-Eval value:
(a b 2)

;;; Amb-Eval input:
try-again

;;; Amb-Eval value:
(a c 3)


What values would have been displayed if we had used set! here
rather than permanent-set!?