Comments by "WhatAboutTheBee" (@WhatAboutTheBee) on "Origins of Precision" video.

  1. At 17:17, you ask for the number of combinations the gauge blocks can form. That is defined by N factorial, denoted "N!". N is the number of blocks. For example, if N=5 blocks, then (N)! = 5×4×3×2×1= 120 combinations. This will over count duplicate lengths but every combination of blocks is accounted for. For example, we can first have .250 and then .125 as one arrangement, except first have .125 then .250 as a second arrangement. There are probably other combinations of blocks which reach .375 as well. Are those to be counted as a unique arrangement, or are you only looking for unique lengths. Let's suppose there are ~50 blocks in that set. 50! = 3 × 10 to the 64th power or 3 followed by 64 zeros. That's just a pinch more than 100,000. Ha! You'd have to write a bit of S/W to find block combinations that have unique lengths. Ditto eliminating different orders of the same block. No matter. The end result will be far and away much greater than 100K.
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