seating arrangement

**kodirl** · August 9th 2006, 03:58 AM

Q.Three girls a,b,c and five boys d,e,f,g,h are seated around a circular table.How many different arrangements are possible if no two girls are to sit together?

**OReilly** · August 9th 2006, 04:27 AM

Originally Posted by kodirl

Q.Three girls a,b,c and five boys d,e,f,g,h are seated around a circular table.How many different arrangements are possible if no two girls are to sit together?

There can be five possible places in table where each girl can seat:

1 d 2 e 3 f 4 g 5 h

Totally possible combinations of a,b,c are 3*2*1 = 6 so there can be totally 6 arrangements of girls in 5 places.

Since each arrangement of girls (for example a,b,c) can be arranged in table in totally 9 combinations around table so that no two girls are together then its totally 6*9 = 45 combinations of girls around table.

Now there can be 5*4*3*2*1 = 120 combinations of boys arrangement.

Multiply total combinations of boys and girls 120*45 and you get that is totally 5400 arrangements if no two girls are siting together.

**ThePerfectHacker** · August 9th 2006, 11:19 AM

Consider the table,

Code:

            1
        8      2
      7           3
        6      4
            5

Now a girl need to take up a seat call that 1.
The other girl cannot sit at 2 or 8.
The possibilities for the other two girls are:

Code:

3 possibilities for girl 1.
And 2*1 possibilities for girls 2 and 3.
For the other 5 positition there are 5!=120 arrangentments for boys.
Thus in total for any seating arragement you have,
$3! \cdot 5!=720$
Times the number of different seatings. Which is 6.
$6\cdot 720=4320$

**Soroban** · August 9th 2006, 02:38 PM

Hello, kodir!

Wow . . . I got yet another answer.

Three girls $a,b,c$ and five boys $d,e,f,g,h$ are seated around a circular table.
How many different arrangements are possible if no two girls are to sit together?

There are two seatings in which no girls are adjacent.

Code:

         G   *             G   *
      *         G       *         G

      *         *       *         *
         *   G             G   *

All others are rotations of these two arrangements.

In both arrangements, the three girls can be seated in $3!$ ways.
. . and the five boys can be seated in $5!$ ways.

Hence, there are: . $2 \times 3! \times 5! \:=\:\boxed{1440\text{ ways.}}$

**ThePerfectHacker** · August 9th 2006, 02:44 PM

Originally Posted by Soroban

Hence, there are: . $2 \times 3! \times 5! \:=\:\boxed{1440\text{ ways.}}$

I agree with you, it seems I did not consider rotations as part of the problem. In that case if you divide by the number of girls by rotations (3) you get 1440 also.

**galactus** · August 9th 2006, 02:50 PM

Yes, it would appear I certainly over-complicated it.

**galactus** · August 9th 2006, 03:46 PM

You know, I was thinkin' , if we only want different order we can use the previous method. Suppose, though, that we wanted to include the arrangements of everyone keeping the same order but in different seats. That is, everyone slides over a chair right or left. That would be 1440*8=11520 arrangements.

Using the 2 arrangements from Soroban's post, we'd have 8*2=16 possible seating arrangements. 16*720=11520.

Whatcha think?.

**OReilly** · August 10th 2006, 02:24 AM

Originally Posted by galactus

You know, I was thinkin' , if we only want different order we can use the previous method. Suppose, though, that we wanted to include the arrangements of everyone keeping the same order but in different seats. That is, everyone slides over a chair right or left. That would be 1440*8=11520 arrangements.

Using the 2 arrangements from Soroban's post, we'd have 8*2=16 possible seating arrangements. 16*720=11520.

Whatcha think?.

I think that (hopefully i am right!) possible seating arrangements is 11*6*120=7920

11 is possible seating arrangements of girls in table.
6 is number of combinations a,b,c.
120 is number of combinations d,e,f,g,h.

**OReilly** · August 10th 2006, 04:22 PM

Originally Posted by galactus

You know, I was thinkin' , if we only want different order we can use the previous method. Suppose, though, that we wanted to include the arrangements of everyone keeping the same order but in different seats. That is, everyone slides over a chair right or left. That would be 1440*8=11520 arrangements.

Using the 2 arrangements from Soroban's post, we'd have 8*2=16 possible seating arrangements. 16*720=11520.

Whatcha think?.

11520 is definitely correct answer.

There is 16 possible seating arrangements around table of a,b,c.
6 is number of combinations a,b,c.
120 is number of combinations d,e,f,g,h.

So, 16*6*120=11520

**ThePerfectHacker** · August 10th 2006, 05:19 PM

Originally Posted by OReilly

11520 is definitely correct answer.

There is 16 possible seating arrangements around table of a,b,c.
6 is number of combinations a,b,c.
120 is number of combinations d,e,f,g,h.

So, 16*6*120=11520

The number different ways to place 8 people at a round table is,
$7!=5040$
So how can you possibly have more with some limitation on the seating!
(Look how well Soroban explained his answer).

**OReilly** · August 10th 2006, 05:41 PM

Originally Posted by ThePerfectHacker

The number different ways to place 8 people at a round table is,
$7!=5040$
So how can you possibly have more with some limitation on the seating!
(Look how well Soroban explained his answer).

These are possible seating arrangements for one combination of girls (for example abc, * are boys):

1: a*b*c***
2: a*b**c**
3: a*b***c*
4: a**b*c**
5: a**b**c*
6: a***b*c*
7: *a*b*c**
8: *a*b**c*
9: *a*b***c
10: *a**b*c*
11: *a**b**c
12: *a***b*c
13: **a*b*c*
14: **a*b**c
15: **a**b*c
16: ***a*b*c

**ThePerfectHacker** · August 10th 2006, 05:49 PM

But some of them are the same. Look at #1 and #16 there is not "head" of the table.

**OReilly** · August 10th 2006, 05:53 PM

Originally Posted by ThePerfectHacker

But some of them are the same. Look at #1 and #16 there is not "head" of the table.

I don't understand. They are not the same. In #1 a,b,c are in 1,3,5 seats and in #16 they are in 4,6,8 seats.

**ThePerfectHacker** · August 11th 2006, 02:22 PM

Although OReilly's method was not entirely correct, it has insipired me to solve this problem with group theory. Let $X$ be the set of all possible arrangements of the seatings from seat #1 to seat #8 (Which is 11520 look at OReilly's post). The problem is that these arrangements are the same if you consider rotations. Let,
$G=\{r_0,r_1,...r_7\}$ be a group of rotations. Define the action of G on X to be $r_i(x)$ is a new arragentment obtained to moving everyone counterclockwise 1 person. Under these definitions we see that X is a G-set. Now, question is finding distinct seatings up to rotations. Which is the same as counting the number of orbits in X under G. Using, the Burnside Formula we have, that that number is,
$\frac{1}{|G|}\sum_{g\in G}|X_g|$ . Note, that $X_g=\{gx=x\}$ can only be the same when $g=r_0 \mbox{ (identity element) }$ . Thus, you have,
$\frac{1}{|G|}\left( |X|+0+0+... \right)$
Which is,
$\frac{1}{8}(11520)=1440$

**OReilly** · August 11th 2006, 03:59 PM

Originally Posted by ThePerfectHacker

Although OReilly's method was not entirely correct, it has insipired me to solve this problem with group theory. Let $X$ be the set of all possible arrangements of the seatings from seat #1 to seat #8 (Which is 11520 look at OReilly's post). The problem is that these arrangements are the same if you consider rotations. Let,
$G=\{r_0,r_1,...r_7\}$ be a group of rotations. Define the action of G on X to be $r_i(x)$ is a new arragentment obtained to moving everyone counterclockwise 1 person. Under these definitions we see that X is a G-set. Now, question is finding distinct seatings up to rotations. Which is the same as counting the number of orbits in X under G. Using, the Burnside Formula we have, that that number is,
$\frac{1}{|G|}\sum_{g\in G}|X_g|$ . Note, that $X_g=\{gx=x\}$ can only be the same when $g=r_0 \mbox{ (identity element) }$ . Thus, you have,
$\frac{1}{|G|}\left( |X|+0+0+... \right)$
Which is,
$\frac{1}{8}(11520)=1440$

My mistake is that I looked at all seating arrangements of girls not distinctive seating arrangements of girls. So from that point of view my answer is correct.
But looking at distinctive seating arrangements in terms how girls can be positioned then my answer is not correct.

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