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Chess with Maths
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You must need to be hold good concepts on probability, permutation and combination then you will slowly realizes why to count 2 cases
How the variation can be more .
I have done my time in the kitchen with all three.
Suppose we are placing a white knight and a black knight on a 2x1 chessboard.
We can place the white knight as below
N .
. N
In each case there is 1 place to put the black knight giving
N n
n N
so you can count the number of cases as 1.1+1.1=2.
Now you want to double that, you say, because there are 2 knights.
That gives 4 cases.
SHOW ME THEM!
Case 1 :When white plays first in 2×1 after black
Fixing white on first ( a1 )
W□>> we have one ways to fill b1 ( let ) --1
Fixing white on b1
□ W >> we have left black place a1 ---2
Now Case 2:- When black plays first then after white
Fixing black on a1
B□ >> we have one ways to fill for white b1
Fixing black on b2
□B>> We have one ways to fill for white on a1
Note that cases got repeated but here sequence matters which we are playing first
There is no connection between the order the pieces are assumed on the board in a method of counting the diagrams and any side to play. There is no mention of side to play in your current enunciation and in the original White to play was specified, so the assumption of separate cases depending on side to play is invalid on any sensible interpretation. You might as well double your cases by counting them as separate depending on whether it's sunny or raining or multiplying by 150 to account for the ply count (actually make that infinity because the positions aren't legal so no reason why the ply count should be).
The salient point is that the final outcomes (diagrams) in your Case 1 and Case 2 are the same, so if you count both you're double counting. A straightforward mathematical error.
You say here , "If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled ." So which is it to be? Are you doubling because there are two queens, or are you doubling because you've decided to have two separate cases for WTP and BTP? If the latter then you need to add it to the question, because it's either been specifically ruled out or not mentioned so far.
That's not mathematical error you should prove it if is so
That's what I just did.
https://youtu.be/Km024eldY1A?feature=shared
Just you see sequence matters in chess board whether to put black aur white queens first .There is total possible cases not only yours case ( which is assumed to be default case only by your mind )
And, as I read it, the minds of almost all of the other contributors, which is not surprising given the wording of your question:
In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place).Cancel means possibilities of capture.
Who would ever read that to mean in how many ways can the queens be placed into the board and a side to move be chosen?
And how would choosing a side to move be described by, "If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled"?
total cases would definitely counted finally be multiplying 2 just because of sequence.
Yes if originally white as to play first then only Cases without multiplying it but now there is both possibilities of black as well as white .
In the original, the comment that White was to play appeared to be irrelevant to the question. I assumed that was why you deleted it.
I deleted because after few more some one will come up with " This is not about playing game " this is about placing queens in that case that might change all prospective.
For a side to move be chosen is default case for all possible ways