19 Replies
Heres what I did
we subtract those cases where knight attacks queen and queen attacks knight from 1 and here there will be no double counting as knight and queen cannot simultaneously attack each other
now consider a 2x3 rectangle
clearly there are 4 possible ways the knight can attack the queen
so no of 2x3 squares in a 8x8 square
7x6
(i did this manually, please let me know if you guys know some other way?)
so number of ways a knight can attack a queen is 7x6x4)
now getting to the no of ways the queen can attack the knight
now for every position of queen on the board there will be 14 squares in same row or column
the main problem is of diagonals
how do i calculate that
for diagonals i thought of doing something like this
like calculate the ways queen can attack in same direction diagonals
but that turns out to be a crap logic as too many cases
how do i calculate diagonals
LEMME THINK TILL TOMORROW IF NOTHING HITS ME I LL ASK MY TEACHER SORRY FOR THE CAPS CAPS LOCK BROKEN
this from todays score i think?
today we didnt have score
alright, i have asked my sir but he hasnt seen the message yet
@Xyphoes
There is a crude but logical method
The sample space is 64P2
Now, assume you're placing the queen in different positions
The knight cannot be present on the row, column or diagonals containing the queen
Then you'll have to eliminate the positions from where the knight can attack the queen (you'll need to take cases here)
Thats what i did
I have written my method, please take a look at that and let me know if i am wrong somewhere
@Apu
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.I did something but now my brain is getting f’ed up
Technically it works
ye to mere paper me aaya tha, but still nothing hit me
nvm i got it
sir ne soln bhej diya
fuck this 1 sec rule i dont even know yeh kyu rakha hai
consider one direction at first
[8c2 + (7c2 + 6c2 + 5c2 + 4c2 + 3c2+2c2)*2]x (2!)
8c2 for the main diagonal
7c2 for the diagonal below it and so on
we multiplied 7c2 6c2 .... by 2 as the same pattern would be observed in the upper half
and at last a 2! as knight and queen can interchange places (permutation)
this is only for one direction
↘️ for this direction same pattern
so multiply the whole thing by 2
you will get 560 as ans
also isme 2 se multiply karenge as
7x6x4 woh cases hai jahan pe 2x3 rectangles horizontal honge hume vertical wale bhi chahiye isliye 2 se multiply
haha cool q. Linar we good?
yes
+solved @Xyphoes
+solved @Xyphoes
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