mathmath333
  • mathmath333
Counting Problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{4 men and 3 women are to be seated in row so that } \hspace{.33em}\\~\\ & \normalsize \text{no two women sit together. } \hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways in which they can be seated. } \hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
ok so we lsit the ways to do this can you think of any
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & a.)\ 4!\times ^{5}P_{3} \hspace{.33em}\\~\\ & b.)\ 4!\times ^{5}C_{3} \hspace{.33em}\\~\\ & c.)\ 4!\times ^{4}P_{3} \hspace{.33em}\\~\\ & d.)\ 4!\times ^{4}C_{3} \hspace{.33em}\\~\\ \end{align}}\)

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More answers

mathmath333
  • mathmath333
@mathmate
mathmath333
  • mathmath333
i have found a link can any one explan mehttp://www.meritnation.com/ask-answer/question/m-men-and-n-women-are-to-be-seated-in-a-row-so-that-no-two-w/permutations-and-combinations/5901983
mathmate
  • mathmate
Yes, it does help!
mathmate
  • mathmate
So we first seat the 4 men, so there are 4! ways of seating the gents. |dw:1440446507784:dw|
dan815
  • dan815
yep that pic explains it perfectly
mathmate
  • mathmate
Now 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
  • dan815
ill give u another question similiar to this , try to figure it out :)
mathmate
  • mathmate
I'll try! lol
dan815
  • dan815
what if the 4 men and 3 women are seated around a round table
mathmate
  • mathmate
with the same constraint?
dan815
  • dan815
yes
mathmate
  • mathmate
lemme see.
mathmath333
  • mathmath333
0 ways
dan815
  • dan815
haha
dan815
  • dan815
ya
mathmate
  • mathmate
It's cheaper!
mathmath333
  • mathmath333
is 0 correct
dan815
  • dan815
ya its impossible ofcourse lol
mathmath333
  • mathmath333
yay!
dan815
  • dan815
|dw:1440447006885:dw|
mathmate
  • mathmate
|dw:1440446969033:dw| So for the four men, there are 3! ways (rotational symmetry)
mathmate
  • mathmate
For the women, there are only 4 seats, so 4P3. Now we multiply them together!
mathmate
  • mathmate
For some reason, men have to fight for women! lol
dan815
  • dan815
remember its not 4 men and 4 women mathmate,
mathmath333
  • mathmath333
yes^ i was wondering that pic
mathmate
  • mathmate
Four seat, but not all filled!
mathmate
  • mathmate
First women gets to choose from 4, then 3, then 2.
mathmate
  • mathmate
*woman
mathmate
  • mathmate
One seat will be empty, so remove the chair and two men will sit next to each other.
dan815
  • dan815
yep
dan815
  • dan815
here is a more involved question
mathmate
  • mathmate
Woah, not too difficult, I hope!
dan815
  • dan815
4 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
  • mathmate
Is this a trick question? They seem like the ones we did.
dan815
  • dan815
no this one is more straightforward but theres some nice ways to do it efficiently
dan815
  • dan815
ill leave u the answers see if u get them both
mathmate
  • mathmate
4!5P4 for the first one, and 3!4P4 for the second.
dan815
  • dan815
part a =1152 part b= 72
mathmate
  • mathmate
So I got them both wrong! lol
mathmath333
  • mathmath333
for first question of round table is this possible|dw:1440447723500:dw|
dan815
  • dan815
oh oops ya, i thought the question said no 2 women and no 2 wmen can sit together
mathmath333
  • mathmath333
i refuted my answer lol
dan815
  • dan815
no i wanted it to be like that, no gender together xD
mathmath333
  • mathmath333
oh it is different
mathmate
  • mathmate
Yea, 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
  • dan815
yep thats right
dan815
  • dan815
ill check your first question answer in the mean time
mathmate
  • mathmate
For 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
  • dan815
u r close
dan815
  • dan815
remember that now since its m and w on a round table mwmwmw = wmwmwm
dan815
  • dan815
so its not 1152/8 but 1152/2/8
mathmate
  • mathmate
Oh, for the straight row, I have 1152=2*4!4! But for the round table, I still have 144=3!4!
dan815
  • dan815
well the way i connected the first answer to the 2nd is
dan815
  • dan815
u see how its 2*4!*4! theres no need for the 2 * anymore in the 2nd case since
dan815
  • dan815
w1m1w2m2 = m1w1m2w2 if u connect the edges its the same arrangement
dan815
  • dan815
so 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
  • dan815
w1m1w2m2 for example this is equal to m2w1m1w2
dan815
  • dan815
|dw:1440448675569:dw|
dan815
  • dan815
so 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
  • mathmate
That'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
  • dan815
yep :)
dan815
  • dan815
do u get it too mathmath333
mathmate
  • mathmate
Sorry, @mathmath333, we hijacked your post! :(
dan815
  • dan815
hehe
mathmath333
  • mathmath333
was offline ,lol
mathmate
  • mathmate
Sorry mm333, we were discussing extensions to your problem! lol

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