All the 7-digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once, and not divisible by 5 All the 7-digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once, and not
divisible by 5, are arranged in the increasing order. Find the 2000-th number in this list.
My try:
The number of 7-digit numbers with 1 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once is 6! = 720.The number of 7-digit numbers with 1 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once is 6! = 720.
 A: Observe the following:


*

*There are $6!-5!=600$ numbers starting with $1$

*There are $6!-5!=600$ numbers starting with $2$

*There are $6!-5!=600$ numbers starting with $3$

*There are $6!-5!=600$ numbers starting with $4$


So your number is the $200$th number which starts with $4$:


*

*There are $5!-4!=96$ numbers starting with $41$

*There are $5!-4!=96$ numbers starting with $42$

*There are $5!-4!=96$ numbers starting with $43$


So your number is the $8$th number which starts with $43$:


*

*There are $4!-3!=18$ numbers starting with $431$


So your number is the $8$th number which starts with $431$:


*

*There are $3!-2!=4$ number starting with $4312$

*There are $3!-2!=4$ number starting with $4315$


So your number is the last number which starts with $4315$.
Therefore your number is $4315762$.
A: The number of 7-digit numbers with 1 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once is 6! = 720. But 120 of these end in 5 and
hence are divisible by 5. Thus the number of 7-digit numbers with 1 in the left most place
and containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once but not divisible by 5 is 600.
Similarly the number of 7-digit numbers with 2 and 3 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once but not divisible by 5 is also 600 each. These
account for 1800 numbers. Hence 2000-th number must have 4 in the left most place.
Again the number of such 7-digit numbers beginning with 41,42 and not divisible by 5 is
120 − 24 = 96 each and these account for 192 numbers. This shows that 2000-th number
in the list must begin with 43.
The next 8 numbers in the list are: 4312567, 4312576, 4312657, 4312756, 4315267, 4315276,
4315627 and 4315672. Thus 2000-th number in the list is 4315672.
Thanks for the hint @almagest
A: 1 at extreme left and rest 6 numbers can be arranged in 6!=720 ways
But 120 numbers end with 5 so 720-120=600
Similarly
Starting with 2, 600 numbers
Starting with 3, 600 numbers
Starting with 41, 96 numbers
Starting with 42, 96 numbers
Starting with 4312, 4 numbers
Total 1996 numbers and
Next 4 numbers are
4315267, 
4315276, 
4315627, 
4315672,  
so 2000th number is 4315672
A: There will be 6 numbers with 4315 _ _ _ instaed of 4 cause you can fill the remaining three places in 3! ways and as 5 is used earlier only then you don't have to neglect 2 case in which 5 comes last
