537471 = [ C(20,10) + C(19,9) + C(18,8) + ... + C(11,1) ]  +  C(20,10)

where C(n,m) = n!/(m!n-m!).  (The extra C(20,10) are the pairs
connecting the subsets into a list.)  I'll leave the proof of this
claim as an exercise for Oleg.

537471 is the number of conses done by the accumulator-based version:

(define (subsets l n)
  (let ss ((l l) (len (length l)) (n n) (little-acc '()) (big-acc '()))
    (cond ((zero? n) (cons little-acc big-acc))
          ((> n len) big-acc)
          (else (ss (cdr l) (- len 1) (- n 1) (cons (car l) little-acc)
                    (ss (cdr l) (- len 1) n little-acc big-acc))))))

You can save an additional whopping ten pairs by reusing parts of the
argument.  I proclaim the following the STINGIEST subsets function
ever known to man:

(define (subsets l n)
  (let ss ((len (length l)) (n n) (little-acc '()) (big-acc '()))
    (cond ((zero? n) (cons little-acc big-acc))
          ((> n len) big-acc)
          (else (let ((last-pair (list-tail l (- len 1))))
                  (ss (- len 1) (- n 1)
                      (if (eq? (cdr last-pair) little-acc)
                          last-pair
                          (cons (car last-pair) little-acc))
                      (ss (- len 1) n little-acc big-acc)))))))

-al