Jorge
Switzerland Lausanne Vaud

Hey there,
I was always interested in how many different combinations/permutations are possible when randomizing all engineers, including the ones from GRR and ME. I have conducted a backoftheenvelope calculations which I am sharing with you! Any feedback or corrections would be highly appreciated.
For the purposes of my calculations, I am considering a 4player setup, with 3 'A' engineers and 4 'B' engineers and the full engineer pool available from both expansions. Nevertheless, I will leave out the coal engineer 'B' numbered #0, as the coal module reduces the number of rounds by one.
"A" engineers
There are 10 engineers:  6 from the base game, numbered #2 to #7.  2 from the GRR, numbered #5 and #6.  2 from the ME, numbered #4 and #7.
* Since we have to select 3 engineers out of 10, we have C(10,3) combinations. * Nevertheless, we have to exclude sets with samenumbered engineers, i.e., (x, x', y). Example: (#5, #5', #3).
There are 4 distinct values that x may take: [47]. There are 8 distinct values that y may take: every other 'A' engineer, i.e., any 'A' except x and x'. Therefore, we have to exclude 4×8 combinations.
Consequently, there are C(10,3)  4×8 = 88 different combinations. Or, multiplied by 3!, there are 528 different permutations.
"B" engineers
This is a little more tricky. There are 4 engineers here, therefore it is possible to have sets like (x,x',y,y').
There are 12 engineers: (barring the #0)  8 from the base game, numbered #8 to #15.  2 from the GRR, numbered #10 and #15.  2 from the ME, numbered #11 and #14.
* Since we have to select 4 engineers out of 12, we have C(12,4) combinations. * Nevertheless, we have to exclude sets with samenumbered engineers, i.e., (x, x', y, y') and (x, x', y, z), with y≠z. Example: (#14, #14', #15, #15') or (#14, #14', #15, #8).
First of all, let's look into these sets: (x, x', y, y') We shall treat this case separately from the subsequent one, so as not to deduct twice some combinations. There are 3 distinct values that x and y may take: #11, #14, #15. All these lead to sets that have to be excluded. Therefore, we have to exclude C(3,2) combinations; allocate 3 values to 2 variables.
Moreover, let's now look into these sets: (x, x', y, z), with y≠z. There are 3 distinct values that x may take: #11, #14, #15. There are 10 distinct values that y may take: every other 'B' engineer, i.e., any 'B' except x and x'. There are 8 distinct values that z may take: every other 'B' engineer except y', i.e., any 'B' except x, x', y and y'. We exclude y' as we have already deducted this set. Therefore, we have to exclude 3×10×8 combinations.
Consequently, there are C(12,4)  C(3,2)  3×10×8 = 252 different combinations. Or, multiplied by 4!, there are 6,048 different permutations.
"A" and "B" combined
Combining the above, we have 88×252 = 22,176 different combinations or 528×6,048 = 3,193,344 different permutations. Thus, there are over 3 million setups!
Base Game This is more straightforward. There are C(6,3) = 20 'A' engineer combinations and C(8,4) = 70 'B' engineer combinations. Multiplying by 3! and 4! respectively, there are 120 'A' engineer permutations and 1,680 'B' engineer permutations. Combined, there are 1,400 different combinations and 201,600 different permutations, which is still impressive!


