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