Because all of the jacks are available as a single plate, it makes sense to
start with those. There are 18 different jacks, each of which can take up the
entire plate.
Single plates: 18
Now we consider both combinations and permutations. In a combination, the order
does not matter. The drawing of random numbers from a pool and getting
'abcd' and 'abdc' are the same combination of letters. If the order matters,
'abcd' and 'abdc' are two different permutations. This means that there will
be many, many more possible permutations than combinations.
In this case, the user will be more discerning than having just 'An HDMI, VGA,
and Cat5e' cable, it's an 'HDMI on the left, VGA in the middle, and Cat5e on the
right' cable, not to be replaced by one where the jacks are in a different
order.
n = number of jacks to choose from
r = how many plates are being filled
combinations = n!
--------
(n-r)!r!
combinations with repetition = (n+r-1)!
--------
r!(n-1)!
permutations = n!
------
(n-r)!
permutations with reptition = n^r
(Recall that n! is a factorial and the result of 1 x 2 x 3 x 4 x ... x n)
Two plates combinations = 153
Two plate combinations with repetition = 171
Two plates permutations = 306
Two plates with repetition = 324
Three plates combinations = 816
Three plate combinations with repetition = 1140
Three plates permutations = 4896
Three plates with repetition = 5832
Four plates combinations = 3060
Four plate combinations with repetition = 5985
Four plates permutations = 73440
Four plates with repetition = 104976
Five plates combinations = 8568
Five plate combinations with repetition = 26334
Five plates permutations = 1028160
Five plates with repetition = 1889568
The sum of all of the permutations of 1, 2, 3, 4, or 5 plates with the above
connectors (one connector per plate) is : 2000718
That's with just one connector per plate!
----
The more complex part is that 15 of the 18 connectors can be put on the top
OR bottom half of a plate (and be duplicated top and bottom) while still being
put side by side up to five times on separate plates
If we go back to the permutation with repetition calculation, we can substitute
15 jacks for r and 2 positions for n, which gives us the total number of
permutations for a two-connector single face plate: 225
Now we add to that total the number of potential whole-face plate configurations
of 18 and get: 243
So there are 243 potential single plates that can be placed in any order up to
five times. Repeating the above calculations we get the following:
Two plates combinations = 29403
Two plate combinations with repetition = 29646
Two plates permutations = 58806
Two plates with repetition = 59049
Three plates combinations = 2362041
Three plate combinations with repetition = 2421090
Three plates permutations = 14172246
Three plates with repetition = 14348907
Four plates combinations = 141722460
Four plate combinations with repetition = 148897035
Four plates permutations = 3401339040
Four plates with repetition = 3486784400
Five plates combinations = 6774333580
Five plate combinations with repetition = 7355513520
Five plates permutations = 812920030000
Five plates with repetition = 847288609000
Summing all of the permutations with repetition (from one to five plates) yields
the grand total of:
850789801599 possible permutations