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?
Originally Posted by kodirl
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.
Consider the table,
Now a girl need to take up a seat call that 1.Code:1 8 2 7 3 6 4 5
The other girl cannot sit at 2 or 8.
The possibilities for the other two girls are:
3 possibilities for girl 1.Code:73 74 75 63 64 53
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,
$\displaystyle 3! \cdot 5!=720$
Times the number of different seatings. Which is 6.
$\displaystyle 6\cdot 720=4320$
Hello, kodir!
Wow . . . I got yet another answer.
Three girls $\displaystyle a,b,c$ and five boys $\displaystyle d,e,f,g,h$ 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 $\displaystyle 3!$ ways.
. . and the five boys can be seated in $\displaystyle 5!$ ways.
Hence, there are: .$\displaystyle 2 \times 3! \times 5! \:=\:\boxed{1440\text{ ways.}}$
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?.
These are possible seating arrangements for one combination of girls (for example abc, * are boys):Originally Posted by ThePerfectHacker
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
Although OReilly's method was not entirely correct, it has insipired me to solve this problem with group theory. Let $\displaystyle X$ 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,
$\displaystyle G=\{r_0,r_1,...r_7\}$ be a group of rotations. Define the action of G on X to be $\displaystyle r_i(x)$ 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,
$\displaystyle \frac{1}{|G|}\sum_{g\in G}|X_g|$. Note, that $\displaystyle X_g=\{gx=x\}$ can only be the same when $\displaystyle g=r_0 \mbox{ (identity element) }$. Thus, you have,
$\displaystyle \frac{1}{|G|}\left( |X|+0+0+... \right)$
Which is,
$\displaystyle \frac{1}{8}(11520)=1440$
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.Originally Posted by ThePerfectHacker
But looking at distinctive seating arrangements in terms how girls can be positioned then my answer is not correct.