Counting Problem

- mathmath333

Counting Problem

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- 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

ok so we lsit the ways to do this can you think of any

- 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

@mathmate

- 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

Yes, it does help!

- mathmate

So we first seat the 4 men, so there are 4! ways of seating the gents.
|dw:1440446507784:dw|

- dan815

yep that pic explains it perfectly

- 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

ill give u another question similiar to this , try to figure it out :)

- mathmate

I'll try! lol

- dan815

what if the 4 men and 3 women are seated around a round table

- mathmate

with the same constraint?

- dan815

yes

- mathmate

lemme see.

- mathmath333

0 ways

- dan815

haha

- dan815

ya

- mathmate

It's cheaper!

- mathmath333

is 0 correct

- dan815

ya its impossible ofcourse lol

- mathmath333

yay!

- dan815

|dw:1440447006885:dw|

- mathmate

|dw:1440446969033:dw|
So for the four men, there are 3! ways (rotational symmetry)

- mathmate

For the women, there are only 4 seats, so 4P3.
Now we multiply them together!

- mathmate

For some reason, men have to fight for women! lol

- dan815

remember its not 4 men and 4 women mathmate,

- mathmath333

yes^ i was wondering that pic

- mathmate

Four seat, but not all filled!

- mathmate

First women gets to choose from 4, then 3, then 2.

- mathmate

*woman

- mathmate

One seat will be empty, so remove the chair and two men will sit next to each other.

- dan815

yep

- dan815

here is a more involved question

- mathmate

Woah, not too difficult, I hope!

- 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

Is this a trick question? They seem like the ones we did.

- dan815

no this one is more straightforward but theres some nice ways to do it efficiently

- dan815

ill leave u the answers see if u get them both

- mathmate

4!5P4 for the first one,
and
3!4P4 for the second.

- dan815

part a =1152
part b= 72

- mathmate

So I got them both wrong! lol

- mathmath333

for first question of round table is this possible|dw:1440447723500:dw|

- dan815

oh oops ya, i thought the question said no 2 women and no 2 wmen can sit together

- mathmath333

i refuted my answer lol

- dan815

no i wanted it to be like that, no gender together xD

- mathmath333

oh it is different

- 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

yep thats right

- dan815

ill check your first question answer in the mean time

- 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

u r close

- dan815

remember that now since its m and w on a round table
mwmwmw = wmwmwm

- dan815

so its not 1152/8
but 1152/2/8

- mathmate

Oh, for the straight row, I have 1152=2*4!4!
But for the round table, I still have 144=3!4!

- dan815

well the way i connected the first answer to the 2nd is

- dan815

u see how its 2*4!*4! theres no need for the 2 * anymore in the 2nd case since

- dan815

w1m1w2m2 = m1w1m2w2
if u connect the edges
its the same arrangement

- 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

w1m1w2m2
for example this is equal to
m2w1m1w2

- dan815

|dw:1440448675569:dw|

- 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

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

yep :)

- dan815

do u get it too mathmath333

- mathmate

Sorry, @mathmath333, we hijacked your post! :(

- dan815

hehe

- mathmath333

was offline ,lol

- mathmate

Sorry mm333, we were discussing extensions to your problem! lol

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