# Math Question, not for HW **URL:** \ **Category:** Factual Questions\ **Created:** [July 16, 2012, 2:24pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411 "2012-07-16T14:24:18Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![pancakes3](https://avatars.discourse-cdn.com/v4/letter/p/ba9def/32.png) [@pancakes3](https://boards.straightdope.com/u/pancakes3)\ **Post date:** [July 16, 2012, 2:24pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/1 "2012-07-16T14:24:18Z")
In how many distinct ways can eight different colored beads be arranged on a circular necklace with no clasp or noticeable break? ---
**Author:** ![Simplicio](https://avatars.discourse-cdn.com/v4/letter/s/c37758/32.png) [@Simplicio](https://boards.straightdope.com/u/Simplicio)\ **Post date:** [July 16, 2012, 2:27pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/2 "2012-07-16T14:27:18Z")
7! ETA: (that’s seven factorial, not me being really excited about the number seven) ---
**Author:** ![Munch](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/munch/32/5281_2.png) [@Munch](https://boards.straightdope.com/u/Munch)\ **Post date:** [July 16, 2012, 2:29pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/3 "2012-07-16T14:29:51Z")
> [@Simplicio](#): > > ETA: (that’s seven factorial, not me being really excited about the number seven) Haha - I was going to say! 7! comes out to 40,320. ---
**Author:** ![fachverwirrt](https://avatars.discourse-cdn.com/v4/letter/f/94ad74/32.png) [@fachverwirrt](https://boards.straightdope.com/u/fachverwirrt)\ **Post date:** [July 16, 2012, 2:35pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/4 "2012-07-16T14:35:51Z")
> [@Munch](#): > > Haha - I was going to say! > > 7! comes out to 40,320. I get 5040. I’m pretty sure you calculated 8! there. ---
**Author:** ![Simplicio](https://avatars.discourse-cdn.com/v4/letter/s/c37758/32.png) [@Simplicio](https://boards.straightdope.com/u/Simplicio)\ **Post date:** [July 16, 2012, 2:38pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/5 "2012-07-16T14:38:09Z")
Its easy to see why if you already know the number of different combinations for n objects where the order matters is n!. On a circular wire, the order still matters, but there’s no “first” spot. But you can choose an arbitrary color to mark the “start” of the series. Then the one to the right is the "first"spot, the one to the right of that the “second” and so on. So you end up with (n-1) spots for (n-1) beads. Thus (n-1)! total combinations. ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [July 16, 2012, 2:45pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/6 "2012-07-16T14:45:33Z")
> [@Simplicio](#): > > Its easy to see why if you already know the number of different combinations for n objects where the order matters is n!. On a circular wire, the order still matters, but there’s no “first” spot. > > But you can choose an arbitrary color to mark the “start” of the series. Then the one to the right is the "first"spot, the one to the right of that the “second” and so on. So you end up with (n-1) spots for (n-1) beads. Thus (n-1)! total combinations. yes, if the following are all equivalent ( necklace with no beginning, no end)… 12345678 23456781 34567812 etc. - 8 different “identical” arrangements. then yes, it’s 8!/8 or 7! ---
**Author:** ![Munch](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/munch/32/5281_2.png) [@Munch](https://boards.straightdope.com/u/Munch)\ **Post date:** [July 16, 2012, 3:35pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/7 "2012-07-16T15:35:27Z")
> [@fachverwirrt](#): > > I get 5040. I’m pretty sure you calculated 8! there. Yup - you’re right. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [July 16, 2012, 3:44pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/8 "2012-07-16T15:44:34Z")
Since the necklace seems to be symmetric, let me … Nitpick: Only 7!/2 2520 arrangements can be obtained from one of the other 2520 arrangements by simply turning the necklace upside-down. ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [July 16, 2012, 4:07pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/9 "2012-07-16T16:07:59Z")
Since there’s no mention that the beads take up the entire necklace, there should be an infinite number of distinct arrangement of beads including the gaps between them. I doubt that’s what the OP had in mind, but I’ve got a lot of experience reviewing customer specs for ambiguities. I wouldn’t sign that contract until it was clarified. On top of that, the OP isn’t clear that there are only 8 beads, just 8 different colors of beads. ---
**Author:** ![yorick73](https://avatars.discourse-cdn.com/v4/letter/y/b5a626/32.png) [@yorick73](https://boards.straightdope.com/u/yorick73)\ **Post date:** [July 16, 2012, 4:36pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/10 "2012-07-16T16:36:53Z")
Wait…there is no break in the necklace so the beads cannot be randomly placed. Wouldn’t that equal just 8 combinations. I think the OP is suggesting that we start with 8 beads already on a closed necklace. Am I missing something basic? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [July 16, 2012, 4:52pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/11 "2012-07-16T16:52:51Z")
The [general solution](https://en.wikipedia.org/wiki/Necklace_%28combinatorics%29) is a little more complicated. You need to specify whether you’re allowing reflections of a necklace to be distinct, and after that you can use the linked formulas. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [July 16, 2012, 5:12pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/12 "2012-07-16T17:12:46Z")
> [@TriPolar](#): > > Since there’s no mention that the beads take up the entire necklace, there should be an infinite number of distinct arrangement of beads including the gaps between them. I doubt that’s what the OP had in mind, but I’ve got a lot of experience reviewing customer specs for ambiguities. I wouldn’t sign that contract until it was clarified. On top of that, **the OP isn’t clear that there are only 8 beads, just 8 different colors of beads**. The OP says “eight different colored beads”. The words “eight”, “different”, and “colored” are all adjectives modifying “beads”, so I don’t see how you’re getting this ambiguity in the number of beads. I will add to the nitpick that we don’t know if the beads are symmetric. If each bead isn’t mirror-symmetric in the along-string direction, and could be strung in either direction, then the number would grow to 7! \* 2^7. ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [July 16, 2012, 5:57pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/13 "2012-07-16T17:57:05Z")
> [@ZenBeam](#): > > The OP says “eight different colored beads”. The words “eight”, “different”, and “colored” are all adjectives modifying “beads”, so I don’t see how you’re getting this ambiguity in the number of beads. > > I will add to the nitpick that we don’t know if the beads are symmetric. If each bead isn’t mirror-symmetric in the along-string direction, and could be strung in either direction, then the number would grow to 7! \* 2^7. In that case the ambiguity is that there may not be eight different colors. We don’t know whether or not ‘eight’ is the extent of the enumeration of ‘color’. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [July 16, 2012, 7:06pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/14 "2012-07-16T19:06:56Z")
> [@yorick73](#): > > Wait…there is no break in the necklace so the beads cannot be randomly placed. Wouldn’t that equal just 8 combinations. I think the OP is suggesting that we start with 8 beads already on a closed necklace. Am I missing something basic? Seconded. I do not understand how to interpret “with no clasp or noticeable break” except as implying that the beads basically _can’t_ be rearranged. If there’s no clasp or noticeable break, how do you get them off the chain to rearrange them? But if that’s not what “no clasp or noticeable break” means, then what _does_ it mean? ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [July 16, 2012, 7:15pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/15 "2012-07-16T19:15:29Z")
> [@septimus](#): > > Since the necklace seems to be symmetric, let me … Nitpick: > > Only 7!/2 > 2520 arrangements can be obtained from one of the other 2520 arrangements by simply turning the necklace upside-down. Doh! Of course, unless they are weird one-side-up-only-can’t-be-flipped beads. (Edit - but then they could be rotated on the string after the necklace is flipped. Doh!) ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 16, 2012, 7:17pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/16 "2012-07-16T19:17:59Z")
> [@Frylock](#): > > Seconded. I do not understand how to interpret “with no clasp or noticeable break” except as implying that the beads basically _can’t_ be rearranged. If there’s no clasp or noticeable break, how do you get them off the chain to rearrange them? > > But if that’s not what “no clasp or noticeable break” means, then what _does_ it mean? It means there’s no distinguished starting spot; rotating the beads around the necklace without changing their adjacency relation does not produce a new arrangement. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [July 16, 2012, 7:28pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/17 "2012-07-16T19:28:49Z")
The necklace is really only a way to visualize the problem. If you’re hung up on the details, think of how many ways there are to arrange the letters in ABCDEFGH if you don’t distinguish between the circular shifts of the words. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 16, 2012, 7:30pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/18 "2012-07-16T19:30:07Z")
In other words, “How many ways can 8 different symbols be arranged into a sequence, where two sequences are considered the same if they are cyclic shifts of each other (the way ABCDEFGH and DEFGHABC are)?”. ETA: Er, I wrote this before refreshing the page; **ultrafilter** ’s post wasn’t there yet. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [July 16, 2012, 9:43pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/19 "2012-07-16T21:43:51Z")
> [@TriPolar](#): > > In that case the ambiguity is that there may not be eight different colors. We don’t know whether or not ‘eight’ is the extent of the enumeration of ‘color’. It doesn’t matter. There’d still be eight different beads, even if some are the same color. Maybe they have different sizes or shapes. ---
**Author:** ![Senegoid](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/senegoid/32/6606_2.png) [@Senegoid](https://boards.straightdope.com/u/Senegoid)\ **Post date:** [July 16, 2012, 10:37pm UTC](https://boards.straightdope.com/t/math-question-not-for-hw/628411/20 "2012-07-16T22:37:13Z")
> [@Simplicio](#): > > 7! > > ETA: (that’s seven factorial, **not me being really excited about the number seven** ) My 11th grade Algebra teacher was quite careful to make the same point, explicitly telling us that 6! (the example he always used), is not pronounced **SIX!** A friend of mine, a molecular biologist and research mathematecian, told me of a time he didn’t know the factorial notation. He saw some derivation or something in his undergrad textbook, which ended up (skipping a few step) pronouncing that the final answer reduced to 6! My friend tried and tried to figure how one got 6 for the answer (not knowing that ! meant factorial), and upon failing to do so, that only reinforced his interpretation that the ! simply meant that this was a surprising and remarkable fact. ! [Next page](https://boards.straightdope.com/t/math-question-not-for-hw/628411.md?page=2)