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Posted - 10/01/2007 :  20:47:21  Show Profile
A row of beads consists of 10 red beads and 5 blue beads.
How do I compute the total number of possible arrangements of the beads that have no two blue beads adjacent?
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2731 Posts

Posted - 10/01/2007 :  21:32:43  Show Profile
The ten red beads create eleven positions into which one can put the blue beads: _R_R_R_R_R_R_R_R_R_R_
Therefore, there is a combination of eleven places taken five at a time to place the blue beads.

Edited by - pka on 10/01/2007 21:34:50
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