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Unformatted text preview: has n2 elements, . .., and the rth has n r elements. The number of distinct ways in which this can be done is n!/n1!n2!. .nr!). Equivalently: Given n objects, n1 of which are of indistinguishable kind, n2 of which are of a second indistinguishable kind, . .., and nr are of rth indistinguishable kind, then the number of distinguishable permutations of the n objects is exactly n!/(n1!n2!. ..nr!)....
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This note was uploaded on 10/14/2007 for the course BTRY 4080 taught by Professor Schwager during the Fall '06 term at Cornell.
- Fall '06