Find the Kth element of a set with arrangements of digits from 1 to n Given a number N between 2 and 9, consider the integers that can be formed using
each digit in the set {1, 2, . . . , N} exactly once. For instance, if N is 3, the integers we
can form are 123, 231, 312, 132, 321 and 213. If we arrange these in increasing order,
we get the list {123, 132, 213, 231, 312, 321}. The fourth number in this list is 231.
In general, given two numbers N and K, the task is to compute the number at position
K when all integers formed using the digits {1, 2, . . . , N} exactly once are arranged in
ascending order. For instance, the example worked out above corresponds to N = 3
and K = 4. 
Find the the answer for the following values of N and K.
 N = 5, K = 76
How am I meant to do this?
 A: You can find an algorithm of linear time like this.
Note that if you order the numbers of $N$ digits in an increasing order, then the first $(N-1)!$ numbers will have $1$ as first digit, the next $(N-1)!$ numbers will have $2$ as a first digit and so on. So determing in which subset of $(N-1)!$ numbers $K$ is determines the first digit. So to determine the first digit find the number:
$$M_1 = \left[\frac{K-1}{(N-1)!} \right]$$
and choose $(M_1+1)$-th number in the set $\{1,2,...,N\}$. Then delete the chosen number from the list and repeat recursively this algorithm while you write all the digits. But this time you have to subtract $M_1 \cdot (N-1)!$ from $K$ and perform the same algorithm with using the new number as $K$. 

In particular let's take a look at the case $N=5$, $K=76$.
We have $M_1 = \left[\frac{76-1}{(5-1)!}\right] = 3$ and then the $4$-th number in $\{1,2,3,4,5\}$ is $4$. Pick it as first digit and then the set of digits is $\{1,2,3,5\}$. 
Now perform again. Assing $76 - 3\cdot 24 = 4$ to $K$ and we have: $M_2 = \left[\frac{4-1}{(4-1)!}\right] = 0$. So pick $1$ as a second digit.
Again. Assign $4 - 0\cdot 6 = 4$ to $K$ and then $M_3 = \left[\frac{4-1}{(3-1)!}\right] = 1$, so pick $3$ as the third digit, as the set of digits is $\{2,3,5\}$ and we need the second smallest.
Repeating this two more times will give you that the $76$-th number in the sorted sequence is $41352$
