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mathmath333
 one year ago
Counting Problem
mathmath333
 one year ago
Counting Problem

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{4 men and 3 women are to be seated in row so that } \hspace{.33em}\\~\\ & \normalsize \text{no two women sit together. } \hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways in which they can be seated. } \hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so we lsit the ways to do this can you think of any

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & a.)\ 4!\times ^{5}P_{3} \hspace{.33em}\\~\\ & b.)\ 4!\times ^{5}C_{3} \hspace{.33em}\\~\\ & c.)\ 4!\times ^{4}P_{3} \hspace{.33em}\\~\\ & d.)\ 4!\times ^{4}C_{3} \hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i have found a link can any one explan me http://www.meritnation.com/askanswer/question/mmenandnwomenaretobeseatedinarowsothatnotwow/permutationsandcombinations/5901983

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3So we first seat the 4 men, so there are 4! ways of seating the gents. dw:1440446507784:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.2yep that pic explains it perfectly

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Now we leave 5= (4+1) chairs between and beside the gents for the three women to choose. So there are 5P3 ways to sit them.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ill give u another question similiar to this , try to figure it out :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2what if the 4 men and 3 women are seated around a round table

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3with the same constraint?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ya its impossible ofcourse lol

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3dw:1440446969033:dw So for the four men, there are 3! ways (rotational symmetry)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3For the women, there are only 4 seats, so 4P3. Now we multiply them together!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3For some reason, men have to fight for women! lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.2remember its not 4 men and 4 women mathmate,

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0yes^ i was wondering that pic

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Four seat, but not all filled!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3First women gets to choose from 4, then 3, then 2.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3One seat will be empty, so remove the chair and two men will sit next to each other.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2here is a more involved question

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Woah, not too difficult, I hope!

dan815
 one year ago
Best ResponseYou've already chosen the best response.24 women and 4 men how many ways to ARRANGE them in a row, where 2 women cannot sit beside each other. How many ways to Arrange them around a round table,(note*the seat number is not what the counts but the pattern of the arrangement*)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Is this a trick question? They seem like the ones we did.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2no this one is more straightforward but theres some nice ways to do it efficiently

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ill leave u the answers see if u get them both

mathmate
 one year ago
Best ResponseYou've already chosen the best response.34!5P4 for the first one, and 3!4P4 for the second.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2part a =1152 part b= 72

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3So I got them both wrong! lol

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0for first question of round table is this possibledw:1440447723500:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh oops ya, i thought the question said no 2 women and no 2 wmen can sit together

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i refuted my answer lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.2no i wanted it to be like that, no gender together xD

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0oh it is different

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Yea, for the first one (4 gents+4 ladies), we put the women on the men's right, so there are 4!4!=576 ways. Now put the ladies on the left of the gents, we get another 576, that makes 1152.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ill check your first question answer in the mean time

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3For the second, I still get 144=3!4!. 3! for the men with rotational symmetry. Once the men are seated, there are 4 distinct seats for the 4 ladies, so 4!

dan815
 one year ago
Best ResponseYou've already chosen the best response.2remember that now since its m and w on a round table mwmwmw = wmwmwm

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so its not 1152/8 but 1152/2/8

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Oh, for the straight row, I have 1152=2*4!4! But for the round table, I still have 144=3!4!

dan815
 one year ago
Best ResponseYou've already chosen the best response.2well the way i connected the first answer to the 2nd is

dan815
 one year ago
Best ResponseYou've already chosen the best response.2u see how its 2*4!*4! theres no need for the 2 * anymore in the 2nd case since

dan815
 one year ago
Best ResponseYou've already chosen the best response.2w1m1w2m2 = m1w1m2w2 if u connect the edges its the same arrangement

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so u have 4!*4! now also since its a round table u can keep shifting every arrangement by 8 times to get the same one around the round table

dan815
 one year ago
Best ResponseYou've already chosen the best response.2w1m1w2m2 for example this is equal to m2w1m1w2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so for 8 seats there are 8 repeated arranements that are coming out of the row case when u move to the round table case

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3That's true, I forgot the men can sit in the women's seats at the very start. I place the men, then throw in the chairs! lol I get it now, 72 is the number!

dan815
 one year ago
Best ResponseYou've already chosen the best response.2do u get it too mathmath333

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Sorry, @mathmath333, we hijacked your post! :(

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Sorry mm333, we were discussing extensions to your problem! lol
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