mathmath333
  • mathmath333
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Find no of ways 7 boys and 7 girls can sit around a round table }\hspace{.33em}\\~\\ & \normalsize \text{ such that no 2 boys or 2 girls are together.}\hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
|dw:1442401794759:dw|
imqwerty
  • imqwerty
:O

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More answers

mathmath333
  • mathmath333
is it 7!*7!
imqwerty
  • imqwerty
|dw:1442401981845:dw|
arindameducationusc
  • arindameducationusc
I guess it is (7-1)!*7! according to the diagram you have given
arindameducationusc
  • arindameducationusc
but the question says 2 boys and 2 girls alternatively?
mathmath333
  • mathmath333
yes in book it is 6!*7!
arindameducationusc
  • arindameducationusc
my bad..... ! I didn't read the question properly, it says no two boys and girls.....
mathmath333
  • mathmath333
how does (7-1)! comes ?
anonymous
  • anonymous
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
mathmath333
  • mathmath333
but by that logic it should be =(7-1)!*(7-1)!=6!*6!
arindameducationusc
  • arindameducationusc
hmmm..... no no
arindameducationusc
  • arindameducationusc
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....
mathmate
  • mathmate
You can think of it this way |dw:1442402619890:dw|
mathmate
  • mathmate
A boy chooses any of the 14 seats, and it won't change anything because of rotation.
mathmate
  • mathmate
Then there are 6! to seat the boys.
mathmate
  • mathmate
The girls sit between the boys, so 7!
anonymous
  • anonymous
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
anonymous
  • anonymous
thats a total of 6! 7!
arindameducationusc
  • arindameducationusc
|dw:1442402759464:dw| just assisting :D
mathmath333
  • mathmath333
ok thnks all
imqwerty
  • imqwerty
no prblm hehe :D
anonymous
  • anonymous
|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
anonymous
  • anonymous
that is 6! * 7! total
anonymous
  • anonymous
|dw:1442403160561:dw|
mathmath333
  • mathmath333
suppose their were 3 persons then ?
anonymous
  • anonymous
|dw:1442403194062:dw|
imqwerty
  • imqwerty
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!
anonymous
  • anonymous
with 3 people, you will have 2 boys or 2 girls sitting next to each other. that is pigeonhole principle
mathmath333
  • mathmath333
|dw:1442403195959:dw|
imqwerty
  • imqwerty
and what are the conditions ? like a cant sit with b or c or something like that?
anonymous
  • anonymous
oh 3 boys and 3 girls
mathmath333
  • mathmath333
no boy, girl ,alien
Jhannybean
  • Jhannybean
alien...lmao
anonymous
  • anonymous
3 boys 3 girls 3 aliens no 2 boys, no 2 girls, no 2 aliens sit next each other?
mathmath333
  • mathmath333
ya this one ^
mathmath333
  • mathmath333
will it be 3!*3!*2!
anonymous
  • anonymous
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
mathmath333
  • mathmath333
3!*3!*2!
anonymous
  • anonymous
|dw:1442403829346:dw|
anonymous
  • anonymous
|dw:1442403873463:dw|
anonymous
  • anonymous
when you start placing girls, you have more choices
mathmath333
  • mathmath333
3!3!6!
anonymous
  • anonymous
|dw:1442403928373:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
or the number of ways to rearrange xxx g1 g2 g3 is 6! / (3! * 1!*1! *1!)
anonymous
  • anonymous
I get 1440
anonymous
  • anonymous
'x' stands for a space a girl will not land in , its blank
mathmath333
  • mathmath333
how u get this "6*5*4".
anonymous
  • anonymous
you saw how i placed the aliens, the boys, then we have left the girls
mathmath333
  • mathmath333
|dw:1442404331340:dw|
mathmath333
  • mathmath333
|dw:1442404352769:dw|
mathmath333
  • mathmath333
|dw:1442404370104:dw|
anonymous
  • anonymous
|dw:1442404420852:dw|
mathmath333
  • mathmath333
what is "x" in ur diagram
anonymous
  • anonymous
those are blank spaces, the girls do not land in them . but they are important in the configuration
anonymous
  • anonymous
the girls can switch places with the x's
mathmath333
  • mathmath333
ok i get it now
anonymous
  • anonymous
then the question is how many ways to rearrange g1 g2 g3 x x x
anonymous
  • anonymous
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
anonymous
  • anonymous
these*
mathmath333
  • mathmath333
yep i agree
mathmath333
  • mathmath333
thats why they are Counting questions
mathmath333
  • mathmath333
|dw:1442404776805:dw|this is imquerty
imqwerty
  • imqwerty
lol
imqwerty
  • imqwerty
how do u make such drawings :D
mathmath333
  • mathmath333
with tool extension
mathmath333
  • mathmath333
works in chrome
imqwerty
  • imqwerty
awww :( i experimented with chrome nd now its not wrkin i need to completely uninstall it but it doesnt wrk ;-;
mathmath333
  • mathmath333
this is that tool ^
imqwerty
  • imqwerty
;-; i will fix it after 2 weeks :( till then i hav to use the old tool ;-;
mathmath333
  • mathmath333
lol it works instantly
anonymous
  • anonymous
this is how math gets deeper and deeper
imqwerty
  • imqwerty
:)
anonymous
  • anonymous
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.
mathmath333
  • mathmath333
this is hard wont come in ma exam
anonymous
  • anonymous
this can be a research problem :)
anonymous
  • anonymous
we have done some base cases
mathmath333
  • mathmath333
yea not for inocents like me
anonymous
  • anonymous
:D
mathmath333
  • mathmath333
|dw:1442405691978:dw|
anonymous
  • anonymous
Thank you for using the OpenStudy Notebook extension.
mathmath333
  • mathmath333
welcome
anonymous
  • anonymous
|dw:1442406626294:dw|
arindameducationusc
  • arindameducationusc
can you guys give me the extension link?
mathmath333
  • mathmath333
The one i attached before will work

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