MHB Another simple propability question

  • Thread starter Thread starter alexmahone
  • Start date Start date
AI Thread Summary
The discussion centers on calculating the probability that no two people occupy adjacent seats when $n$ people are seated in $2n$ seats. The initial solution incorrectly states the number of favorable arrangements as $2(n!)^2$, while the correct formula is $(n+1)(n!)^2$. The total arrangements for seating $n$ people in $2n$ seats is accurately given as $\frac{(2n)!}{n!}$. A clarification is provided by suggesting that testing with specific values, such as $n=2$, can help identify the error in reasoning. The correct probability formula is confirmed to be $\frac{(n+1)(n!)^2}{(2n)!}$.
alexmahone
Messages
303
Reaction score
0
If $n$ people are seated in a random manner in a row containing $2n$ seats, what is the probability that no two people will occupy adjacent seats?

My solution:

There are $2n!$ favourable arrangements.

Total no. of arrangements:

The 1st person can sit on any of the $2n$ seats, the 2nd person can sit on any of the other $2n-1$ seats and so on. So, the number of ways to seat $n$ people on $2n$ seats is $\displaystyle 2n(2n-1)\cdots (n+1)=\frac{(2n)!}{n!}$.

$\displaystyle P=\frac{2(n!)^2}{(2n)!}$

But the answer given in the book is $\displaystyle P=\frac{(n+1)(n!)^2}{(2n)!}$. Where have I gone wrong?
 
Mathematics news on Phys.org
Alexmahone said:
If $n$ people are seated in a random manner in a row containing $2n$ seats, what is the probability that no two people will occupy adjacent seats?

My solution:

There are $2n!$ favourable arrangements.

Total no. of arrangements:

The 1st person can sit on any of the $2n$ seats, the 2nd person can sit on any of the other $2n-1$ seats and so on. So, the number of ways to seat $n$ people on $2n$ seats is $\displaystyle 2n(2n-1)\cdots (n+1)=\frac{(2n)!}{n!}$.

$\displaystyle P=\frac{2(n!)^2}{(2n)!}$

But the answer given in the book is $\displaystyle P=\frac{(n+1)(n!)^2}{(2n)!}$. Where have I gone wrong?

The number of ways to place $n$ people on $2n$ seats such that no two sit adjacent to each other is not $2(n!)$. Its equal to $(n+1)(n!)$.
Take $n=2$ and list out the possible ways to seat 2 people on 4 seats. You will realize your mistake.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top