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[HomieDolphine]

The prize isn't available anymore. DP of the tokens will be at 15:00 today (server time)
the answer: The final answer is (2N)!, which is equal to 1*2*3*...*(2N-1)*2N. the explanation: Firstly decide which guests get the white hats. we have 2N guests and N white hats, so there are 2NcN ways to do this, which is ((2N)!/(N!)^2). Now in order to define the circles, we will define two functions which are onto and one to one between the group of people with white hat and the group of people with black hat (make sure you understand why defining those functions gives all the possible circles arrangements and every arrangement is given exactly one time). The first function will be for every person with white hat to decide excactly which person with black hat will stand to his right, there are N! ways to define this function. The second function will be for every person with black hat the decide exactly which person with white hat will stand to this right, there are N! ways to define this function. so we have ((2N)!/(N!)^2)*(N!)^2=(2N)!. Final answer: there are (2N)! ways.

 

[HomieDolphine]

n²-1? acc to me but idk idk

Sorry but this is a wrong answer

 

[NinetyBac]

Let's say that there are people Named 1,2,3,4...
And let us assume all people with odd numbered name gets a white hat and all the people with even numbered name gets a black hat,
And now let's arrange them in a line with all white hatted people adjacent to all the black hatted people,
And it is clear that the circle must contain even number of people as it would be impossible for two people with same coloured hat not standing near each other,
By this, for the first arrangement, John shall pair all the people adjacent to each other and they form n circles
The possible ways of arranging people such that every circle has two members is N! as in the first person in the white line could be paired with n people in the black line and the second person in the white line could be paired with N-1 persons in black line as the forst person already got a pair, and the third person: N-2 ans the Nth person would be 1

This is the for the case where there is two people in each circle

For the rest for the cases where there are more than two people in each circle,
Let's say one circle contains 4 people and the rest contains 2 people...

My brain cells hurt to be calculating this, so this is as far as I've gotten. Gl to everyone who attempts this

*ALSO STOP CALLING ME A NERD*

 

[NinetyBac]

Let's say that there are people Named 1,2,3,4...
And let us assume all people with odd numbered name gets a white hat and all the people with even numbered name gets a black hat,
And now let's arrange them in a line with all white hatted people adjacent to all the black hatted people,
And it is clear that the circle must contain even number of people as it would be impossible for two people with same coloured hat not standing near each other,
By this, for the first arrangement, John shall pair all the people adjacent to each other and they form n circles
The possible ways of arranging people such that every circle has two members is N! as in the first person in the white line could be paired with n people in the black line and the second person in the white line could be paired with N-1 persons in black line as the forst person already got a pair, and the third person: N-2 ans the Nth person would be 1

This is the for the case where there is two people in each circle

For the rest for the cases where there are more than two people in each circle,
Let's say one circle contains 4 people and the rest contains 2 people...

My brain cells hurt to be calculating this, so this is as far as I've gotten. Gl to everyone who attempts this

*ALSO STOP CALLING ME A NERD*

 

[HomieDolphine]

Let's say that there are people Named 1,2,3,4...
And let us assume all people with odd numbered name gets a white hat and all the people with even numbered name gets a black hat,
And now let's arrange them in a line with all white hatted people adjacent to all the black hatted people,
And it is clear that the circle must contain even number of people as it would be impossible for two people with same coloured hat not standing near each other,
By this, for the first arrangement, John shall pair all the people adjacent to each other and they form n circles
The possible ways of arranging people such that every circle has two members is N! as in the first person in the white line could be paired with n people in the black line and the second person in the white line could be paired with N-1 persons in black line as the forst person already got a pair, and the third person: N-2 ans the Nth person would be 1

This is the for the case where there is two people in each circle

For the rest for the cases where there are more than two people in each circle,
Let's say one circle contains 4 people and the rest contains 2 people...

My brain cells hurt to be calculating this, so this is as far as I've gotten. Gl to everyone who attempts this

*ALSO STOP CALLING ME A NERD*

You made a really good progress, as a mathematics maniac (and a self proclaimed nerd) this is exactly the thing i want to encourage, thinking math.
However your answer for the case with only 2 circles is almost correct. You forgot that while there are exactly N! ways to arrange these people to dance in two circles, you need to multiply by 2N Choose N, or (2N)!/((N!)^2), in order to count all of the arrangments for the hats on the heads of the people. for example if there are 2 people exactly, so there N=1, and N!=1. but there are 2 arrangments. the other arrangment is when you switch the hat colors between the people.
So the answer for only 2 people in a circle case is (2N)!/(N!), or (N+1)*(N+2)*...*(2N)

 

[HomieDolphine]

just gonna assure you that you're not gonna find an someone that can solve imo problems playing on a cracked minecraft server

It's not from IMO, I saw WAY harder problems there. This is actually considered a little bit easy among IMO problems (but not the easiest)

 

[NinetyBac]

You made a really good progress, as a mathematics maniac (and a self proclaimed nerd) this is exactly the thing i want to encourage, thinking math.
However your answer for the case with only 2 circles is almost correct. You forgot that while there are exactly N! ways to arrange these people to dance in two circles, you need to multiply by 2N Choose N, or (2N)!/((N!)^2), in order to count all of the arrangments for the hats on the heads of the people. for example if there are 2 people exactly, so there N=1, and N!=1. but there are 2 arrangments. the other arrangment is when you switch the hat colors between the people.
So the answer for only 2 people in a circle case is (2N)!/(N!), or (N+1)*(N+2)*...*(2N)

Oh right, I've failed to consider the possibility of switching the hat colors

 

[SSSnipS]

Ok i tried solving this but as soon as I seen the factorials I have now decided to officially give up

 

[htsdfisy]

lets just say N is 2 it means there are 4 guests and maximum of 2 circles then there can be 2 arrangements. then if N is 3 then there can be a maximum of 3 circles then there can be 3 arrangements

so the answer is N?

i know im completely wrong but i tried for fun

 

[MrBeastAlan]

I think htsdfisy is correct H HomieDolphine

 

[MeOwHow]

i mean, there is an n number of guests so you're saying that there are the same number of circles as guests so i dont think the ans is correct, but then again i am not the greatest at math.