# Permutations Involving Places

• Jan 17th 2011, 11:53 AM
rebghb
Permutations Involving Places
Hello Everyone!

Lets say we have 6 places: $A_1, A_2, B_1, B_2, C_1, C_2$ and we want to pass by these places in a way that we always start by $A_1$ and end by $A_2$ and that $X_1$ always comes after $X_2$. I'm thinking of $4! / 4 = 6$, I know it's 6, why I used this formula is because it works for 4 places: $A_1, B_1, A_2, B_2$...

(1) What if each time, our trip starts by $B$ or $C$, we get a total of $3*6=18$?

(2) What about when there are 8 places, is it $6!/6 = 120$?

Thanks for the help!
• Jan 17th 2011, 12:11 PM
awkward
Quote:

Originally Posted by rebghb
Hello Everyone!

Lets say we have 6 places: $A_1, A_2, B_1, B_2, C_1, C_2$ and we want to pass by these places in a way that we always start by $A_1$ and end by $A_2$ and that $X_1$ always comes after $X_2$. I'm thinking of $4! / 4 = 6$, I know it's 6, why I used this formula is because it works for 4 places: $A_1, B_1, A_2, B_2$...

We can disregard $A_1$ and $A_2$ because they are fixed at the beginning and end, respectively. There are 4! ways to arrange the Bs and Cs without any restrictions; in 1/2 of these $B_1$ comes before $B_2$, and in half of the ones with $B_1$ before $B_2$ we also have $C_1$ before $C_2$. So the total number of arrangements is

$\frac{1}{2} \cdot \frac{1}{2} \cdot 4! = 6$.

Quote:

(1) What if each time, our trip starts by $B$ or $C$, we get a total of $3*6=18$?
I'm not sure what you mean. If you start with a B, do you have to end with a B? And do you have to start with $B_1$?

Quote:

(2) What about when there are 8 places, is it $6!/6 = 120$?

Thanks for the help!
No, $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 6!$.
• Jan 17th 2011, 12:25 PM
rebghb
Quote:

I'm not sure what you mean. If you start with a B, do you have to end with a B? And do you have to start with http://www.mathhelpforum.com/math-he...ce57e6ebe1.png?

Yes, I mean B1 ... B2 and C1 ... C2

Regards,
• Jan 17th 2011, 01:22 PM
Plato
It has taken me quite sometime to begin to understand this question.
If $n\in \mathbb{Z}^+$ and we have $n$ pairs so there are $2n$ places.
If we begin and end a string with members of the same pair, we can choose that pair in $n$ ways.
The other $(n-1)$ pairs can be arranged in $\dfrac{(2n-2)!}{2^{n-1}}$

So the total ways is $n\cdot\dfrac{(2n-2)!}{2^{n-1}}$

Have I understood what you meant?
• Jan 17th 2011, 01:31 PM
rebghb
Thank you