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combinatorics [2013/04/07 21:27]
jongbor
combinatorics [2014/04/19 15:41] (current)
Line 101: Line 101:
 The answer for distinguishable people and distinguishable groups is simply $k^n$, because of the rule of product: the first person has $k$ buses to choose from, the second person also has $k$ choices, etc. The answer for distinguishable people and distinguishable groups is simply $k^n$, because of the rule of product: the first person has $k$ buses to choose from, the second person also has $k$ choices, etc.
  
 +What if we require each group to have at least one member? (I'm not sure, but I thing it might be related to the Stirling number of the second kind plus some shuffling. I'm not writing a definite formula here since I haven'​t proved it).
 ==== Indistinguishable people and distinguishable groups ==== ==== Indistinguishable people and distinguishable groups ====