Let's see who is smart enough to solve this combinatorial

do your own homework brainlet

>>8848357
Is this a stars-and-bars problem?

>>8848357
This problem is horribly underdefined. So without more information, there is no "solution."

100%

>>8848357
100% if all 20 seats are open

>>8848357
18/19 + 16/17 + 14/15 + 12/13 + 10/11

File: de8.png (8KB, 430x443px) Image search: [Google]

8KB, 430x443px

>>8848447
>probability larger than 1

>>8848444
>>8848431

I don't get it, they can literally sit one next to other

>>8848449
forgot to divide the sum by 20

>>8848450
>men
>invading other men's personal space

>>8848457
>>8848357
he must be gay

>>8848460
Pr(OP=Faggot) = 1.0

>>8848357
>wants people to do math
What the fuck do you come to this board for?

>We decode empty chairs with $0$s and non-empty chairs with $1$s and consider binary strings of length $20$ with $15$ zeros and $5$ ones. Since no men are allowed to sit next to each other we are asking for words which do not contain a string $11$.

>The number of all binary strings of length $20$ with $15$ zeros is
\begin{align*}
\binom{20}{15}=15504
\end{align*}

In order to count the number of strings of length $20$ without having a substring $11$ we consider words
with no consecutive equal characters at all. These words are called Smirnov words or Carlitz words. (See example III.24 *Smirnov words* from *[Analytic Combinatorics](http://algo.inria.fr/flajolet/Publications/books.html)* by Philippe Flajolet and Robert Sedgewick for more information.)

>A generating function for the number of Smirnov words over a binary alphabet is given by
\begin{align*}
\left(1-\frac{2z}{1+z}\right)^{-1}\tag{1}
\end{align*}

We replace occurrences of $0$ in a Smirnov word by one or more zeros since there are no restrictions to them. This corresponds to a substitution of

\begin{align*}
z\longrightarrow z+z^2+\cdots=\frac{z}{1-z}\tag{2}
\end{align*}

Since we want to look for strings of length $20$ with precisely $5$ ones, we mark them with $t$

\begin{align*}
z\longrightarrow tz\tag{3}
\end{align*}

>We obtain by substituting (2) and (3) in (1) a generating function A(z,t)
\begin{align*}
A(z,t)&=\left(1-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}-\frac{tz}{1+tz}\right)^{-1}\\
&=\frac{1+tz}{1-z-tz^2}
\end{align*}

>To obtain the number of words of length $20$ with $5$ ones we calculate with some help of Wolfram Alpha
\begin{align*}
[z^{20}t^5]A(z,t)=4368
\end{align*}
and conclude the probability that $5$ man won't sit next to each other in a row of $20$ chairs is
\begin{align*}
\frac{4368}{15504}\simeq 0.2817
\end{align*}

>>8848457
/thread

Pure autism

>>8848357
You're mother.

>>8848357
If they are engineers the amount of seats taken while all 5 are sitting is 1. That should help you out OP

>>8848357
100%
all the people are women or fembois

>>8848357
it's 1 - probability that 5 men DO sit next to eachother in a row of 20 chairs

>>8848357
>analysis, algebra, and topology fags btfo by a simple counting problem

And they think they "understand" how infinite sets work. Cracks me up.

>>8849015
MY SIDES ARE IN ORBIT