The Antisocial Club Challenge

The Antisocial Club meets every week at Jim's Bar. Since they are so antisocial, however, everyone always sits as far as possible from the other members, and no one ever sits right next to another member. Because of this, the 25-stool bar is almost always less than half full and unfortunately for Jim the members that don't sit at the bar don't order any drinks. 

Jim, however, is pretty smart and makes up a new rule: The first person to sit at the bar has to sit at one of two particular stools. If this happens, then the maximum number of members will sit at the bar. 

Which stools must be chosen? Assume the stools are numbered 1 to 25 and are arranged in a straight line.

Wise Selection of first 2 Seats made available! 

Logical Response to The Antisocial Club Challenge

What was the challenge?

The bar owner Jim put only 2 options for first person to seat and those options are Stool No. 9 and Stool No.17. We'll call seat for stool for convenience.

Choice of numbers 9 and 17 is because of symmetry explained below.

Suppose the first person choose seat no.9.

Then the second person has to choose the furthest seat i.e. seat no.25.

The third person will choose seat no.1 which is furthest from seat no.9 and 25.

The fourth person won't have option other than seat no.25 which is the furthest from rest of 3.

For the fifth person, seats numbered 5, 13 and 21 are available which are equidistant from 2 persons. We assume he chooses farthest from the seat occupied by person who entered just ahead of him. That is he chooses seat no.5.

Since, seat no.21 is furthest from seat no.5 and only seat equidistant from 17 & 25, the sixth person will choose seat no.21.

With that, 7th person won't have any option other than seat no.13 to maintain distance from others.

With the same logic, next 6 persons occupies seat numbered 3, 23, 7, 19, 11, 15.

So there will be 13 people maximum in the bar with no one seating next to each other.

The same seat numbers will be occupied even if the first person choose seat no.17. The second person will choose seat no.1, third will choose seat no.25, fourth will choose seat no.9 and so on.

But if the first person doesn't choose 9 or 17 then there can be less than or equal to 12 people maximum in the bar following their antisocial trait.

Count The Number of Arrangements

There are 10 parking spaces numbered from 101 to 110. At least one car is parked in these slots. If cars can be parked only at the consecutively numbered parking slots, how many such arrangements can be made. 

Consider that only one car can be parked in one parking slot and all cars are identical.

Here is the possible count! 

Possible Number of Arrangements

What was the puzzle?

Suppose there is only 1 car that is to be parked in 1 of the 10 slots. 

Number of possible arrangement = 10C1 = 10!/1!9 = 10.

That is 1 car can be parked in 10 slots in 10 number of ways.

Now, let's suppose that there are 2 cars that to be parked in 2 of the 10 parking slots. But the condition is that they need to be parked in consecutive slots. 

Among 10 slots for there are 9 possible consecutive slots for 2 cars. That is, 2 cars can be parked in consecutive slots in 9C1 = 9 number of ways. It's like placing 1 group of cars (having 2 cars) in 9 possible slots.

Similarly, in 10 parking slots for parking 3 cars there are 8 possible consecutive slots. Hence, there are 8 such arrangements are possible.

And so on for the rest number of cars.

Hence, there are total 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 such arrangements are possible.  

The Unfair Arrangement!

Andy and Bill are traveling when they meet Carl. Andy has 5 loaves of bread and Bill has 3; Carl has none and asks to share theirs, promising to pay them 8 gold pieces when they reach the next town.

They agree and divide the bread equally among them. When they reach the next town, Carl offers 5 gold pieces to Andy and 3 to Bill.

“Excuse me,” says Andy. “That’s not equitable.” He proposes another arrangement, which, on consideration, Bill and Carl agree is correct and fair.

How do they divide the 8 gold pieces?

This is fair arrangement of gold distribution! 

Source 

Correcting The Unfair Arrangement!

How unfair the arrangement was?

First we need to know how 8 loaves (5 of Andy & 3 of Bill) are equally distributes among 3.

If each of them is cut into 2 parts then total 16 loaves would be there which can't be divided equally among 3.

Suppose, each of loaves is divided into 3 parts making total 24 loaves available.

Now, Andy makes 15 pieces of his 5 loaves. He eats 8 and gives the remaining 7 to Carl.

Bill makes 9 pieces of his 3 loaves. He eats 8 and gives the remaining 1 to Carl.

This way, Carl too gets 8 pieces and 8 breads are distributed equally among 3.

 
Obviously, Carl should pay 7 gold pieces to Andy for his 7 pieces and 1 gold piece to Bill for the only piece offered by Bill.