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Just a note

Once you work through the mathematics it turns out that, if we ignore the order in which they are achieved, there are [2n-1]! / [n!(n-1)!] possible ways of selecting n items (with replacement) from a group of n, distinguishable, items.
  • For example, if you have a sample of 2 categorical observations, of value a and b, there are 2×2=4 ways in which you could select 2 observations from that sample: a then a, a then b, b then a, or b then b. Ignoring the order of selection, there are just 3 combinations: a&a, a&b or b&b.
  • If you have a sample of 3 observations, a, b and c, there are 3×3×3 = 33 = 27 ways of selecting 3 observations from that sample. Ignoring the order of selection, there are just 10 combinations: a&a&a, a&a&b, a&a&c, b&b&a, b&b&b, b&b&c, c&c&a, c&c&b, c&c&c, or a&b&c.
  • There are 1010=10,000,000,000 ways of selecting 10 observations from 10 unique observations, and ignoring the order of selection, there are 92,378 combinations.
  • There are 2020 ≅ 1.05×1026 ways of selecting 20 from 20 unique observations, but ignoring the order of selection, there are about 6.9×1010 combinations.