This question is based on permutation concept ? Am I correct ?
Chess with Maths
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Well in the old Scots math means good, so...
Mar 14, 2024
0
#5
I think so, there are two queens and 64 squares , According to me the answer would be 2^8 .
Ashvath23 wrote:
This question is based on permutation concept ? Am I correct ?
Yes
Ashvath23 wrote:
I think so, there are two queens and 64 squares , According to me the answer would be 2^8 .
No
Suppose you have placed your queen on a1 then for another queen to arrange position no neither a square nor 1 square file and the diagonal square would be left for another queen
You have counted all the cases
Mar 14, 2024
0
#8
So, the better way to solve this would be , to left either queen untouched and move another queen , so this will happen in every square on the board , such that it cover all the square and positions .
Mar 14, 2024
0
#9
We can observe that in some cases the numbers are increasing and decreasing .
Hello
Nice to see you after long time...
Mar 14, 2024
0
#11
I've always tried to do this on a chessboard but never could!
At max I could put 7 queens not attacking each other, 8 is very hard.
Yet, very surprised that there are 42 ways 🤯
Optimissed wrote:
There are 42 ways the Qs can be placed with the white Q on a1. There are 42 ways if the white Q is on a4. If the white Q is on f4, there are 38. Logically, for all the edge squares there are 42. If on b7 there are 40, so a pattern is developing. If we divide the board into concentric tracks, the outer track contains squares all of which are 42. The next track inwards 40, the next track inwards, 38.
The final track is the centre four squares. Taking one at random, d4, there are 5x6 +6 = 36
So we have 4x36 + 12x38 + 20x40 + 28x42 = 2576
Double it and we have the answer.
5152
Yes correct and you logically proved 👏
Congrats
Wind wrote:
I've always tried to do this on a chessboard but never could!
At max I could put 7 queens not attacking each other, 8 is very hard.
Yet, very surprised that there are 42 ways 🤯
I think you misunderstood the question , there are only two queens on board
Ashvath23 wrote:
We can observe that in some cases the numbers are increasing and decreasing .
Yes
Isn't it just a multiplication problem?
(64 - (8*3))*64 = 2560?...
Oh... I changed my mind...
Just out of curiosity, what does "never get cancel by each other" mean in the puzzle? Everyone seems to understand it, but I don't.
(64 - (8*3))*7 = 280
(64 - (8*3) -((2*1+3*2+4*3+5*2+6*1)/9))*9 = 324
(280 + 324)*4 = 2416?
🎯Brain teaser ‼️
Can you answer this question with logic?