9 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.Where am I going wrong
1. We don't need to subtract cases like 708 or 609
2. The condition of number must be greater than 212 means cases like 177 are void I think.
i am not subtracting those
3c1 makes any one of the box value equal to 10
if its the middle box the number would look like 2 10 2 , with 1 coin yet to be distributed
this makes me realise my error
will post a new soln shortly
@SirLancelotDuLac thoughts?
If you are comfortable with multinomial, this is coeffecient of x¹⁵ in (x²+x³+...+x⁹)(1+x+x²+...+x⁹)² ig.
There could be a better method though ig...
(x²+x³+...+x⁹)^2 (1+x+x²+...+x⁹)nahi hoga ??