Let n and k be fixed positive integers of the same parity, k >= n.
We are given 2n lamps numbered 1 through 2n; each of them can be on or off.
At the beginning all lamps are off. We consider sequences of k steps.
At each step one of the lamps is switched (from off to on or from on to off).
* Let N be the number of k-step sequences ending in the state:
  lamps 1, ..., n on, lamps n+1, ..., 2n off.
* Let M be the number of k-step sequences leading to the same state
  and not touching lamps n+1, ..., 2n at all.
Find the ratio N/M.

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It makes sense that k,n must be of the same parity, otherwise,
there is no sequence of switches -- the parity of the lamps turned on
in the end equals the parity of the number of steps (= k).

Experiments:
n = 1, k = 1 -> one sequence
... in general, if k = n, the ratio N/M is one because every sequence
    turning on the lamps 1, ..., n have to touch each of these lamps
    exactly once and does not have time to touch the lamps n+1, ..., 2n
    By the way, M = N = n!
n = 1, k = 3:
* N: (1,1,1), (1,2,2), (2,1,2), (2,2,1)
  * only one of the sequences counts for M: (1,1,1)
* N/M = 4
n = 1, k = 5:
* 11111, 22221, 21221, ...
  * in general, if n = 1,
    then M = 1 (only swapping lamp1) and
    N = 2^(k-1) because we can choose any first k-1 steps,
    and then we are forced to choose a given lamp at the end
    by parity.
n = 2, k = 4
* M: 1112, 1121, 1211, 2111, 2221, 2212, 2122, 1222: 8 sequences
* N-M: 12xx for x=3,4, and 4*3 possible positions of 1,2, that is
       24 sequences
* N = 32
* N/M = 4
Hm...
Maybe N/M is always 2^(k-n). (it fits the experiments so far)

How to prove it...
I have an M-sequence, and I would produce 2^(k-n) N-sequences out of it.
Or maybe, I can produce 2^k objects out of an M-sequence, and 2^n objects out of an N-sequence
  so that the objects are unique, and they are the same in both cases.

N-sequence, 1...2^n -> ?
* 2^n represents a subset of {1,2,...,n}
* So the combination is something like initial state of the first n lamps, and the sequence
  * or perhaps the subset of {1,2,...,n} encodes some swapping of the lamp i with i+n?

M-sequence, 1...2^k -> ?
* 2^k represents a subset of {1,2,...,k}, that is the subset of steps
* I could alter the steps of the subset in a way, let's say change
  i -> i+n.
  * So for example sequence (1,2,1,1) + subset(0,1,0,1) -> sequence (1,4,1,3)
  * What is the property of such a sequence? In the end, the lamps (3,4) are lighting.
  * In general, after such modification of the sequence, we always get a sequence in which
    exactly one of the lamps (i, i+n) is on because the parity of lamps which are on
    among (i, i+n) remains the same as in the original sequence.
  * This is promissing.

So the destination object is a k-sequence such that after performing the steps,
for every i = 1...n, exactly one lamp of (i,i+n) is on, let's call it D-sequence.
We have already described a bijection (M-sequence, subset of {1...k}) -> D-sequence.
* (the reverse mapping is obvious, just subtract n from every step touching a lamp > n)

Now, the bijection (N-sequence, subset of {1...n}) -> D-sequence.
* For every i in the subset, make a replacement in the sequence i <-> i+n at every position
  * Or at the first position? Maybe it could also work but I like every position more now.
  * So for example: N-sequence (4,2,1,4), subset {2} -> (2,4,1,2)
    * It apparently leads to a D-sequence.
    * Reversely, given a D-sequence, we look at which lamps are on in the end of the D-sequence
      from that, we read the subset of {1...,n}, and perform the reverse replacement.

QED, N/M = 2^(k-n)

Problem Solved!