In how many ways can 5 beads with different colours be sticked together by a circular wire?

I thought in something round I must do like (5-1)!=4!=24, but the answer is 12..

Printable View

- Jul 3rd 2013, 06:37 AMTrefoil2727permutation question
In how many ways can 5 beads with different colours be sticked together by a circular wire?

I thought in something round I must do like (5-1)!=4!=24, but the answer is 12.. - Jul 3rd 2013, 10:49 AMShadowKnight8702Re: permutation question
I have always had trouble with circle permutations myself. Normally we would simply use (n-1)!. However, since this is a circular wire that can move around in space, you over count the permutations by 2. For problems like this, the required formula is (n-1)!/2. If you want an even better understanding, I would suggest researching the origins of the formula. With this formula, you can divide your 24 by 2 and get 12.

- Jul 3rd 2013, 10:50 AMebainesRe: permutation question
You can think it through like this:

If the question was about 5 beads on a straight wire the answer qwould be 5!. Now when you join the neds of the wire into a loop you divide by 5 to eliminate the duplicates that you get by simply rotating the loop: for example the arrangment A-B-C-D-E is the same as B-C-D-E-A by simply rotating all the beads one position clockwise. I suspect that's how you got to 4! as your answer. But there is one more trick - for any arrangement you get an alternate mirror-image arrangement by__turning the loop over__. Thus A-B-C-D-E is the same as E-D-C-B-A., and so you have to divide the answer of 4! by 2, giving 12. - Jul 3rd 2013, 11:39 AMSorobanRe: permutation question
Hello, Trefoil2727!

Your answer is almost correct.

But this problem has a twist that you haven't recognized.

Quote:

In how many ways can 5 beads with different colours be placed on a circular wire?

I thought in something round I must do like: .(5 - 1)! = 4! = 24, but the answer is 12.

ebaines is absolutely correct.

If we are placing 5 distinct beads in a circle (on a table),

. . the answer is $\displaystyle 4! = 24.$

But we are placing the beads*on a circular wire.*

. . That is, we are creating "bracelets".

$\displaystyle \text{So that: }\;\;\begin{array}{ccccc} &&A \\ \\ E &&*&&B \\ \\ &D&&C \end{array}\;\;\text{ is the same as }\;\;\begin{array}{ccccc}&&A \\ \\ B &&*&& E \\ \\ &C &&D \end{array}$

To eliminate duplication by*reflection*, we divide by 2.

. . There are:.$\displaystyle \frac{24}{2} \,=\,12$ possible bracelets.