In layman's terms, what is a multifactorial? - $x\underbrace{!!\cdots!}_{n\text{ times} }$ My first impression was that given a multifactorial expression, one was to factorial the first term, then factorial that term, then factorial that term, etc. etc.
So, 20!! = (20!)!
I now understand that 20!! = 20(20 - 2)(20-4)....
and 20!!! = 20(20 - 2)(20 - 4)(20 - 6)...
but how would you put that in terms that your average US high-school algebra student might understand?
 A: The double factorial is reasonably standard and has the meaning you state.  $n!!$ is the product of all the numbers down from $n$ that are of the same parity.  I think your statement that $20!!=20(20-2)(20-4)\dots (20-18)$ captures it within the range of a high school student.  You might include an example where $n$ is odd.  The triple factorial is much less common, but if people understand the double factorial the triple is not much of a stretch.  We recently had a question where $a!!$ meant $(a!)!$ and $a!!!$ meant $((a!)!)!$ so you need some care.
A: The factorial is $n! = n(n-1)(n-2)\cdots 2\cdot 1$
The double factoral is $n!! = n (n-2)(n-4) \cdots $ (terminates with $2$ or $1$)
The triple factorial is $n!!! = n(n-3)(n-6)\cdots $ (terminates with $3,2,$ or $1$)
And so forth.  $n\underbrace{!!\ldots!!}_{m} = \prod_{k=0}^{\lceil n/m-1\rceil} (n-km)$ 
Once the distinction between $(n!)!$ and $n!!$ is stressed (it is a unfortunately confusing notation), the average Australian high school student readily grasps how it works.
Getting anyone to remember the notation is another matter.
A: $n!!$ is the product of positive integers $\le n$ in steps of $2,$
$n!!!$ is the product of positive integers $\le n$ in steps of $3,$ and so on
A: To solve a multifactorial, we can start by counting the number of ! we need to write to represent the factorial. We'll call this number "k". We'll call the number we are to factorialize "x".
Now, we set up a sequence where a(n) = x - kn.
Our next step is to take all a(n) from n=1 up to, but not including, the nearest rounded-down integer to n = x/k, and multiply them together and by x.
So, let's take the double factorial of 6. This is a double factorial, and we'll need to write 2 exclamation points to represent this number. So a(n) = 6 - 2n. We now find all a(n) from n = 1 to n = 2.
a(1) = 6 - 2 = 4
a(2) = 6 - 2(2) = 6 - 4 = 2
Now, let's multiply that whole thing together and by 6:
6 * a(1) * a(2) = 6 * 4 * 2 = 12 * 4 = 48.
Alternatively
If you like, you can also think of a(n) as x - 2(n - 1) or x - 2n + 2. In that case, all that you'd then need to do is multiply all a(n) from n = 1 to n = (x - 2)/2. For 6!!,
a(1) = 6 - 2 + 2 = 6
a(2) = 6 - 4 + 2 = 6 - 2 = 4
a(3) = 6 - 6 + 2 = 2
and 6!! simply becomes a(1) * a(2) * a(3) = 6 * 4 * 2 = 48.
