Help pls

The smallest integer on the list is 12345

Second smallest integer is 12354

The first 24 smallest integers is 1---- (since 4!=24)

The next 24 smallest integers are (2----) (again, since 4!=24)

The next 24 smallest integers are (3-----) (since 4!=24)

However, I have already counted 72 integers. The question asks for the 60th integer. So, instead of the ten-thousand's place, I will go for the thousand's place. 

I already know that the integer is between 25431 (the largest integer of the list that starts with 2, this is the 48th smallest integer)

and 35421 (the largest integer that starts with 3 this is the 72nd smallest integer)

I can further narrow it down , since I know that 31--- is 6 integers (since 3!=6), and so that's 48+6=54 integers. 

I know that 32--- is another 6 integers (again, since 3!=6), so once again, that's another 6 integers. 

32541 is the largest integer in the list of the form 32---. So, 32541 is the 60th integer of the list. 

 

JP

See the wiki page
https://en.wikipedia.org/wiki/Factorial\_number\_system
for information on the Factorial Number System.
First, we write the base-10 number 60 in the Factorial
number system.  We calculate
 

    $60 \div 1 = 60$, remainder 0

    $60 \div 2 = 30$, remainder 0

    $30 \div 3 = 10$, remainder 0

    $10 \div 4 = 2$, remainder 2

    $2 \div 5 = 0$, remainder 2

This gives $60_{10} = 2:2:0:0:0_!$
Thus, the 60th permutation of 1, 2, 3, 4, 5 in lexicographic order
is $23145$.