Hi Guest, you have not done it the lazy way, you have done it the most straight forward (and long) way.
I'll see if I get the same answer...
All the 720 arrangements of the digits 1 through 6 are put in numerical order. What is the 428th term?
How many are between 100 000 and 199 000 That would be 1*5! = 120
200 000 - 299 000 That would also be 1*5! = 120
300 000 - 399 000 That would also be 1*5! = 120
So that is 360 so far 428-360= 68 more...
So it is going to be between 400,000 and 499,999
How many between 410,000 and 419,999
the position of the 4 and the 1 is set so that would be 4! = 24 68-24= 44 more to go
How many between 420,000 and 429,999. Another 24 44-24=20 more to go
So it is going to be between 430,000 and 439,999
How many between 431,000 and 431,999. 3!=6
How many between 432,000 and 432,999. 3!=6
We canot repeat the 3 or the 4 at the front so
How many between 435,000 and 435,999. 3!=6 20-18=2 more to go
Next is 436 125, then 436,215
So I think 436,215 is the 428th term. That is what I get
check
120+120+120+24+24+6+6+6+2 = 428 Seems ok
they are not the same so one of us has made a mistake