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The maximum number of combinations for a Latin Hypercube of <math>M</math> divisions and <math>N</math> variables (i.e., dimensions) can be computed with the following formula:
<math>(\prod_{n=0}^{M-1} (M-n))^{N-1} = (M!)^{N-1}</math>
For example, a Latin hypercube of <math>M = 4</math> divisions with <math>N = 2</math> variables (i.e., a square) will have 24 possible combinations. A Latin hypercube of <math>M = 4</math> divisions with <math>N = 3</math> variables (i.e., a cube) will have 576 possible combinations.
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