R999
chetan2u,
BunuelI was trying to solve this question by 324721 to 458521 .. where last 2 digits are fixed.
From 324721 to 399999 -- we can have only 8 (2 to 9) *6(4 to 9) * 3 (7 to 9) = 144 numbers ( as 3 and last 2 digits are fixed).
Similarly from 400000 to 458521 .. we can have 6(0 to 5 for ten thousands digit) * 9 * 6=324 ways.
so total 144+324 =468 ways.
Although i understood the answer for the question given... i did not understand through my above approach where am i goiing wrong. could you please help with this
Hi...
You miss out on a lot of combinations this way..
Just one example..
From 324721 to 399999 -- we can have only 8 (2 to 9) *6(4 to 9) * 3 (7 to 9) = 144 numbers ( as 3 and last 2 digits are fixed).
So you have taken hundreds place as only 7 to 9..
But if you have a number 325121.. this is a valid number but in your approach is not valid because of you taken hundreds as only 7, 8 or 9..
You could this by taking..
324700 to 324999....1*1*1*3*1*1=3
325000 to 329999....1*1*5*10*1*1=50
330000 to 399999..1* 7*10*10*1*1=700
400000 to 449999...1*5*10*10*1*1=500
450000 to 457999....1*1*8*10*1*1=80
458000 to 458600.. 1*1*1*6*1*1=6
Total ... 3+50+700+500+80+6=1339
A