# Thread: Confused on Algebra 2 counting principles story problem

1. ## Confused on Algebra 2 counting principles story problem

This is it:

A Person is wearing a bracelet with 12 settings around the bracelet. How many different ways can 12 birthstones be arranged around the bracelet? Leave the answer in factorial form.

Thanks

2. You have twelve stones to go into twelve settings. How many different ways can you arrange them? If we replace settings with boxes, does it make more sense to you?

Either way, the answer is: $\displaystyle 12!$

3. Originally Posted by colby2152
You have twelve stones to go into twelve settings. How many different ways can you arrange them? If we replace settings with boxes, does it make more sense to you?

Either way, the answer is: $\displaystyle 12!$

Thanks for helping
Just i geuss that is wrong, its what i thought just i tried to submit it as an answer (im doing algebra 2 on a computer based program) and it is wrong the answer they were looking for is $\displaystyle 11!$. The problem is i have no idea why

4. Originally Posted by Audriella
Thanks for helping
Just i geuss that is wrong, its what i thought just i tried to submit it as an answer (im doing algebra 2 on a computer based program) and it is wrong the answer they were looking for is $\displaystyle 11!$. The problem is i have no idea why
The Factorial Rule states that for n different items, there are n! arrangements.

5. The answer is $\displaystyle 11!$. Remember that this is a bracelet, not a string.
You can place n things around a circle in $\displaystyle (n-1)!$ ways.

6. this is super late because i had to go somewhere but...

oh!!! I get it i swear you just add in a formula that relates and things always begin to click with math.

Thanks for the help.

7. Originally Posted by wingless
The answer is $\displaystyle 11!$. Remember that this is a bracelet, not a string.
You can place n things around a circle in $\displaystyle (n-1)!$ ways.
Wingless, that is a good point. In that case, the 12 rotations around the bracelet of $\displaystyle 11!$ arrangements is removed. IMO, the question was lacking details, but this is surely what the creators of the question were going for.

8. For those acquainted with Permutations and Combinations,
. . but unfamiliar with the "Circular Table Problem",
. . this is a classic "trick question."

There are six people to be seated in six chairs set around a circular table.
In how many ways can they be seated?

We must understand that a rotation of an arrangement is not a different arrangement.
. . These two seating are considered equivalent.
Code:
       A       B           F       A
* - - *             * - - *
/       \           /       \
F*         *C       E*         *B
\       /           \       /
* - - *             * - - *
E       D           D       C

The reasoning is like this . . .

The first person can sit anywhere . . . it doesn't matter.
. . Then the other five can be seated in $\displaystyle 5! = 120$ ways.

With a bracelet, an additional "trick" is involved.

Once again, rotations are not counted.
In addition, reflections are also not counted.
. . These two bracelets are considered equivalent.
Code:
       A       B           B       A
* - - *             * - - *
/       \           /       \
F*         *C      C *         *F
\       /           \       /
* - - *             * - - *
E       D           D       E

With twelve (distinguishable) beads on a bracelet,
. . there are: .$\displaystyle 11!$ circular arrangements.

And we must eliminate reflections.
. . There are: .$\displaystyle \frac{11!}{2} \:=\:19,958,\!400$ possible bracelets.

9. Originally Posted by Soroban

With twelve (distinguishable) beads on a bracelet,
. . there are: .$\displaystyle 11!$ circular arrangements.

And we must eliminate reflections.
. . There are: .$\displaystyle \frac{11!}{2} \:=\:19,958,\!400$ possible bracelets.
I thought that too. But the question says, 'A Person is wearing a bracelet'. You can't change the sides of the bracelet while you're wearing it, can you? ^^

10. Originally Posted by wingless
I thought that too. But the question says, 'A Person is wearing a bracelet'. You can't change the sides of the bracelet while you're wearing it, can you? ^^
Good point, wingless. I think you are right
I always hated such counting problems, since they have multiple interpretations. And half the time I have to guess what the teacher is thinking. Anyway I always ended up getting wrong answers.
But the wearing part was neat