14 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.ABCDE mein se koi 3 choose kare (5c3) and 3 rows mein rakhida (3!) rows mein alag alag boxes mein rkh skte so 332 . 5 spaces and 2 letters left so 5c2 * 2!
5c3 * 3! * 3 *2 *2 * 5c2 * 2!
where am i going wrong
another approach is total - non favourable
8c5 - ( 5c5 + 5c5 + 6c5) 5! = 5760
yea this makes more sense tbh
why dis wrong
Well one error seems to be it doesn't account for the places where the initial three are kept, also it can overcount if you overcome the previous as if A is chosen in 5c3 step and B chosen in 5c2 step can replicate a case where b chosen in 5c3 step and a chosen in 5c2 step.
1. it does . the 3 * 3 * 2 part
2. kinda saw this coming . the 5c3 part refers to choosing 3 letters out of 5 whereas the 5c2 part referes to choosing 2 spaces out of 5
Well the thing for the second part remains the same. If a is chosen in 5c3 part and is placed at the first row first column cell and b is placed by remaining 5c2 elements
On the cell just right to that
And the opposite
hmm i see the overcounting now
@hardcoreisdead
yahan pe colors ulte hai but you get the idea right
yeahh
+solved @professional_procrastinator @Lance who?
Lol
mb
lol