উদ্ভাস-উন্মেষ শিক্ষা পরিবার · YouTube · 10.8 MB
· 154s
একটি ৮×৮ দাবার বোর্ডে ৩টি Rook বা নৌকা এমনভাবে বসাতে হবে, যেন কোনো দুটি Rook একই সারি কিংবা কলামে না থাকে। বলো তো, কাজটি মোট কতভাবে করা সম্ভব?
সমাধানটি বুঝতে প্রথমে ১×১, ২×২ ও ৩×৩ দাবার বোর্ডে Rook বসানোর প্যাটার্ন দেখানো হয়েছে। এরপর Combination ও Factorial ব্যবহার করে বের করা হয়েছে ৮×৮ বোর্ডের উত্
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Antor, you want to stop me? This is not good.
Checkmate.
This rook in game is always going to be a problem. Antur, I'll give you an interesting problem.
At the end by end our boarded
How can you keep these three rules in a way that no one will be affected?
That's okay.
I know the route, but I don't know the way.
Column can go as far as the body can go. And if he wants, he can go as far as the body can go. So, the problem you gave, it's not a solution.
Antur, you should think about it. I have a one-by-one application.
Then I can keep one ingredient in one way.
Now you can think, I want to use a 2x2.
We can talk about both of them there.
We can keep it total.
Do it again.
Now, let's say I have a question.
And how many of these three options can I keep three holes?
Hit that hotel.
Try that.
It didn't actually.
You can see that it is following a pattern. There is a space in one factory in one by one database.
2 by 2 daba bode.
There are two holes in 2-factor layout and in 3x3 mode
3 factors are kept in the key factor
So, if I had 8 shapes in this 8x8 board, I could have kept 8 fractal shapes.
So if we want to keep these three groups,
That's all I'm going to talk about today.
In an 8x8 scale, I have 8 squares and 8 columns.
I can choose from 8 to 3.
and from 8 columns, I can choose 3 columns
It's a three-way.
Now, this three sarees and three columns make a three by three board.
In this 3x3 board,
3 is in the 3 factorial. So the final answer is 8c3 * 8c3 * 3 factorial.
So, if our vote is N by M accuracy, then here I can keep R as MCR into MCR into R factorial so that no one attacks anyone.
Now think about it. How many bishops can be kept in a single place so that no one attacks them?
Good luck to all of you who took the problem from the bishop and gave the problem from the knight. The knight's move was not good enough to calculate.
But the fun part
But the knight's case is easier than the bishop's.
Actually, I won't solve this. This is from your side.