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mathmath333
 one year ago
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.
mathmath333
 one year ago
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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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{Find no of ways 7 boys and 7 girls can sit around a round table }\hspace{.33em}\\~\\ & \normalsize \text{ such that no 2 boys or 2 girls are together.}\hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442401794759:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2dw:1442401981845:dw

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1I guess it is (71)!*7! according to the diagram you have given

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1but the question says 2 boys and 2 girls alternatively?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1yes in book it is 6!*7!

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1my bad..... ! I didn't read the question properly, it says no two boys and girls.....

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1how does (71)! comes ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you have n items, the number of ways to rearrange them in a circle is (n1)! you have 14/2 pairs of boy girl , so there are 6! ways to arrange them

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but by that logic it should be =(71)!*(71)!=6!*6!

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1hmmm..... no no

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1Its 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
 one year ago
Best ResponseYou've already chosen the best response.4You can think of it this way dw:1442402619890:dw

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4A boy chooses any of the 14 seats, and it won't change anything because of rotation.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Then there are 6! to seat the boys.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4The girls sit between the boys, so 7!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first 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
 one year ago
Best ResponseYou've already chosen the best response.0thats a total of 6! 7!

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442402759464:dw just assisting :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442403041001:dwfor each arrangement of boys, the girls can rearrange in 7! ways. and the boys can rearrange in 6! ways on a circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is 6! * 7! total

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442403160561:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1suppose their were 3 persons then ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442403194062:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2ok 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
 one year ago
Best ResponseYou've already chosen the best response.0with 3 people, you will have 2 boys or 2 girls sitting next to each other. that is pigeonhole principle

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442403195959:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2and what are the conditions ? like a cant sit with b or c or something like that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh 3 boys and 3 girls

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1no boy, girl ,alien

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03 boys 3 girls 3 aliens no 2 boys, no 2 girls, no 2 aliens sit next each other?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1will it be 3!*3!*2!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lets 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442403829346:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442403873463:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when you start placing girls, you have more choices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442403928373:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0remember 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
 one year ago
Best ResponseYou've already chosen the best response.0or the number of ways to rearrange xxx g1 g2 g3 is 6! / (3! * 1!*1! *1!)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0'x' stands for a space a girl will not land in , its blank

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1how u get this "6*5*4".

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you saw how i placed the aliens, the boys, then we have left the girls

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442404331340:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442404352769:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442404370104:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442404420852:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1what is "x" in ur diagram

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0those are blank spaces, the girls do not land in them . but they are important in the configuration

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the girls can switch places with the x's

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then the question is how many ways to rearrange g1 g2 g3 x x x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what 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

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1thats why they are Counting questions

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442404776805:dwthis is imquerty

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2how do u make such drawings :D

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1with tool extension

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2awww :( i experimented with chrome nd now its not wrkin i need to completely uninstall it but it doesnt wrk ;;

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1this is that tool ^

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2;; i will fix it after 2 weeks :( till then i hav to use the old tool ;;

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1lol it works instantly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is how math gets deeper and deeper

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how 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
 one year ago
Best ResponseYou've already chosen the best response.1this is hard wont come in ma exam

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this can be a research problem :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we have done some base cases

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1yea not for inocents like me

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442405691978:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for using the OpenStudy Notebook extension.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442406626294:dw

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1can you guys give me the extension link?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1The one i attached before will work
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