Math Goodies is a free math help portal for students, teachers, and parents.
|
Interactive Math Goodies Software

testing left nav
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
 All Forums  Homework Help Forums  Miscellaneous Math Topics  Combinations New Topic  Topic Locked  Printer Friendly
Author  Topic

dasims2k5
New Member

USA
1 Posts

 Posted - 10/01/2007 :  20:47:21 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?

pka

USA
2731 Posts

 Posted - 10/01/2007 :  21:32:43 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
Topic
 New Topic  Topic Locked  Printer Friendly Jump To: Select Forum New Visitor Forum       Testing Forum Homework Help Forums       Basic Math and Pre-Algebra       Algebra       Geometry and Trigonometry       Pre-Calculus and Calculus       Probability and Statistics       Standardized Test Preparation Help       Miscellaneous Math Topics Educator Forum       Teacher Talk Parent Forum       Parent's Place  -------------------- Home Active Topics Frequently Asked Questions Member Information Search Page
 Math Forums @ Math Goodies © 2000-2004 Snitz Communications