Find no of ways 7 boys and 7 girls can sit around a round table such that no 2 boys or 2 girls are together.

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Find no of ways 7 boys and 7 girls can sit around a round table such that no 2 boys or 2 girls are together.

Mathematics
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\(\large \color{black}{\begin{align} & \normalsize \text{Find no of ways 7 boys and 7 girls can sit around a round table }\hspace{.33em}\\~\\ & \normalsize \text{ such that no 2 boys or 2 girls are together.}\hspace{.33em}\\~\\ \end{align}}\)
|dw:1442401794759:dw|
:O

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Other answers:

is it 7!*7!
|dw:1442401981845:dw|
I guess it is (7-1)!*7! according to the diagram you have given
but the question says 2 boys and 2 girls alternatively?
yes in book it is 6!*7!
my bad..... ! I didn't read the question properly, it says no two boys and girls.....
how does (7-1)! comes ?
if you have n items, the number of ways to rearrange them in a circle is (n-1)! you have 14/2 pairs of boy girl , so there are 6! ways to arrange them
but by that logic it should be =(7-1)!*(7-1)!=6!*6!
hmmm..... no no
Its because the arrangement for the people anticlockwise and clockwise will be same hence these arrangements are considered as one because order is not changing..... hmmmm..... LET me think of a better explanation....
You can think of it this way |dw:1442402619890:dw|
A boy chooses any of the 14 seats, and it won't change anything because of rotation.
Then there are 6! to seat the boys.
The girls sit between the boys, so 7!
first count the number of ways to rearrange pairs of bg in a circle bg , bg, bg, bg, bg , bg, bg this is 6! now for each of these circle arrangements, you can also rearrange the girls , fixing the boys. there are 7! ways to rearrange the girls
thats a total of 6! 7!
|dw:1442402759464:dw| just assisting :D
ok thnks all
no prblm hehe :D
|dw:1442403041001:dw|for each arrangement of boys, the girls can rearrange in 7! ways. and the boys can rearrange in 6! ways on a circle
that is 6! * 7! total
|dw:1442403160561:dw|
suppose their were 3 persons then ?
|dw:1442403194062:dw|
ok see this :) - |dw:1442403062721:dw| fix the boys nd then u can arrange the grls in 7! ways now u move the boys nd then u will have 6! ways to rearrange we don't count 7! cause we already counted 1 in when we fixed the boys :) so total ways = 6!*7!
with 3 people, you will have 2 boys or 2 girls sitting next to each other. that is pigeonhole principle
|dw:1442403195959:dw|
and what are the conditions ? like a cant sit with b or c or something like that?
oh 3 boys and 3 girls
no boy, girl ,alien
alien...lmao
3 boys 3 girls 3 aliens no 2 boys, no 2 girls, no 2 aliens sit next each other?
ya this one ^
will it be 3!*3!*2!
lets label them b1, g1, a1 b2, g2, a2 b3, g3, a3 place them on a circle fix the aliens with a thumbtack, and start counting the b's and g's
3!*3!*2!
|dw:1442403829346:dw|
|dw:1442403873463:dw|
when you start placing girls, you have more choices
3!3!6!
|dw:1442403928373:dw|
remember we started by fixing a1,a2,a3 the aliens with a thumbtack there are 2! ways to rearrange those. for each alien there are 3! ways to place the boys. for each boy there are 6*5*4 ways to place the girls 2! * 3! * 6*5*4
or the number of ways to rearrange xxx g1 g2 g3 is 6! / (3! * 1!*1! *1!)
I get 1440
'x' stands for a space a girl will not land in , its blank
how u get this "6*5*4".
you saw how i placed the aliens, the boys, then we have left the girls
|dw:1442404331340:dw|
|dw:1442404352769:dw|
|dw:1442404370104:dw|
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what is "x" in ur diagram
those are blank spaces, the girls do not land in them . but they are important in the configuration
the girls can switch places with the x's
ok i get it now
then the question is how many ways to rearrange g1 g2 g3 x x x
what is interesting about theses problems, it is the first group that we 'fix' , (imagine placing thumbtacks on them), that have the circular permutations. the rest are regular or non circular permutations
these*
yep i agree
thats why they are Counting questions
|dw:1442404776805:dw|this is imquerty
lol
how do u make such drawings :D
with tool extension
works in chrome
awww :( i experimented with chrome nd now its not wrkin i need to completely uninstall it but it doesnt wrk ;-;
this is that tool ^
;-; i will fix it after 2 weeks :( till then i hav to use the old tool ;-;
lol it works instantly
this is how math gets deeper and deeper
:)
how to solve the general problem. suppose we have x11 , x12, x13, ... x1k cannot sit next to each other x21, x22, x23 ... , x2k cannot sit next to each other ... ... xn1, xn2, xn3, ... xnk cannot sit next to each other on a circle, how many ways can they be seated.
this is hard wont come in ma exam
this can be a research problem :)
we have done some base cases
yea not for inocents like me
:D
|dw:1442405691978:dw|
Thank you for using the OpenStudy Notebook extension.
welcome
|dw:1442406626294:dw|
can you guys give me the extension link?
The one i attached before will work

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