October 25, 2008

mid-semester

Filed under: Math — Jonathan @ 12:04 am

I have been so busy the past few weeks I haven’t had time for anything! Things have calmed down a bit for the time being, but I know it won’t last too long! I’m just heading to bed but I wanted to at least keep things interesting here and post a math problem! A little combinitorics, so have fun with it! Solution is in the comments!

A bracelet is made by threading 4 identical red beads and 4 identical yellow
beads onto a hoop. How many different arrangements of the 8 beads can be made?

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1 Comment »

  1. Note that a bracelet can be rotated or flipped over without changing the basic pattern.
    Here is one way to count the possibilities: place all four red beads. There are now four gaps in which
    we can place the yellow beads. 1) Place one in each gap. 2) Place all four in one gap. 3 & 4) Place
    one in one gap and three in another – either adjacent or opposite. 5 & 6) Place two in each of two
    gaps – either opposite or adjacent. 7 & 8) Place two in one gap and one in each of two other
    gaps. The two and the empty gap can be either opposite or adjacent.

    Comment by cookla — October 25, 2008 @ 12:04 am


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