1. ## Diner Seating

(Sorry not sure if this is basic or university level)

A diner has 16 two-seat booths.

26 guys walk in at random.

what is the probability that at least one booth is empty?

...

Is this right:

Total seating arrangements = P( 32, 26) = 3.655 * 10^32

...

Arrangements with 3 empty booths = 26! * P( 16, 3 ) = 1.355 * 10^30

Arrangements with 2 empty = P( 28, 26) * P( 16, 2 ) = 3.659 * 10^31

Arrangements with 1 empty = P( 30, 26) * 16 = 1.768 * 10^32

...

(seating arrangements with empty booths) / (total seating arrangements)

= ( 2.148 * 10^32 ) / ( 3.655 * 10^32 ) = 0.588

...

2. Originally Posted by billym
(Sorry not sure if this is basic or university level)

A diner has 16 two-seat booths.

26 guys walk in at random.

what is the probability that at least one booth is empty?

...

Is this right:

Total seating arrangements = P( 32, 26) = 3.655 * 10^32

...

Arrangements with 3 empty booths = 26! * P( 16, 3 ) = 1.355 * 10^30

Arrangements with 2 empty = P( 28, 26) * P( 16, 2 ) = 3.659 * 10^31

Arrangements with 1 empty = P( 30, 26) * 16 = 1.768 * 10^32

...

(seating arrangements with empty booths) / (total seating arrangements)

= ( 2.148 * 10^32 ) / ( 3.655 * 10^32 ) = 0.588

...

Hi billym,

I think you have some problems with over-counting. Your arrangements with 1 empty booth have been included in the count of 2 empty booths, etc.

Are you familiar with the Principle of Inclusion / Exclusion, also known as the Sieve Principle? This might be a good place to apply it.

3. Originally Posted by awkward
Hi billym,

I think you have some problems with over-counting. Your arrangements with 1 empty booth have been included in the count of 2 empty booths, etc.

Are you familiar with the Principle of Inclusion / Exclusion, also known as the Sieve Principle? This might be a good place to apply it.
Okay so I was forgetting to subtract when then 2 empty seats circumstance yielded an event when the guys arranged themselves in a way that left another empty seat. And in the 1 empty seat circumstance there would be the events of both one extra empty seat and two extra empty seats that would need to be subtracted. (???)

So: (16!/13!)*26! + (30!/4!) + (28!/2!)[(16!/14!) - 28)] = 4.472 * 10^31

(4.472 * 10^31) / (3.655 * 10^32) = 0.122

Does that look more reasonable? Am I doing anything right?

4. Can somebody please show me how to do this problem? I've tried it a billion ways and I keep getting a different answer. All it says is "focus on the empty seats."

5. Originally Posted by billym
Can somebody please show me how to do this problem? I've tried it a billion ways and I keep getting a different answer. All it says is "focus on the empty seats."
Let’s reframe the problem. “How many ways can we put twenty-six identical balls into sixteen different boxes if the boxes can not hold more two balls?”
Let d stand for double occupancy, s stand for single occupancy and e stand for empty.
Now we have four cases;
$\displaystyle \begin{array}{ccc} d & s & e \\ \hline {13} & 0 & 3 \\ {12} & 2 & 2 \\ {11} & 4 & 1 \\ {10} & 6 & 0 \\ \end{array}$.
To find the number of ways place the balls is the sum of those four cases.
For the second case: “how many ways can we arrange a string of 12 d’s, 2 s’s and 2 e’s?”
We notice that only one of those cases do we have no empty box.

6. My brain is so fried right now i only kind of understand. I get the chart bit. Is it a simple as the probability of having at least one empty box is 0.75?

What is the purpose of "For the second case: “how many ways can we arrange a string of 12 d’s, 2 s’s and 2 e’s?”

Could you actually write down the correct answer so I'll know when I've done it right?

7. The total number of ways to fill those boxes is $\displaystyle 41328$.
The total number of ways to get the last row is $\displaystyle 8008$
So the probability of having no empty booth is $\displaystyle \frac{{8008}}{{41328}} = 0.194$.
The to the question is $\displaystyle 1-0.194=0.806$

It will help if you go to this website Wolfram|Alpha
In the input window, type in this exact expression: expand (1+x+x^2)^16 (you can copy & paste)
Click the equals bar at the right-hand end of the input window.
In the expansion you will see the coefficient of $\displaystyle x^{26}$ is $\displaystyle 41328$ that is no accident.

See the coefficient of $\displaystyle x^{26}$ is $\displaystyle 8008$ that is also no accident.