Counting Problem

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\(\large \color{black}{\begin{align} & \normalsize \text{4 men and 3 women are to be seated in row so that } \hspace{.33em}\\~\\ & \normalsize \text{no two women sit together. } \hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways in which they can be seated. } \hspace{.33em}\\~\\ \end{align}}\)
ok so we lsit the ways to do this can you think of any
\(\large \color{black}{\begin{align} & a.)\ 4!\times ^{5}P_{3} \hspace{.33em}\\~\\ & b.)\ 4!\times ^{5}C_{3} \hspace{.33em}\\~\\ & c.)\ 4!\times ^{4}P_{3} \hspace{.33em}\\~\\ & d.)\ 4!\times ^{4}C_{3} \hspace{.33em}\\~\\ \end{align}}\)

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Other answers:

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

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