Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$, at least one of the first $k$ terms of the permutation is greater than $k$.
If the first number is $6$, then there are no restrictions. There are $5!$, or $120$ ways to place the other $5$ numbers.

If the first number is $5$, $6$ can go in four places, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.

If the first number is $4$, ....
4 6 _ _ _ _ $\implies$ 24 ways
4 _ 6 _ _ _ $\implies$ 24 ways
4 _ _ 6 _ _ $\implies$ 24 ways
4 _ _ _ 6 _ $\implies$ 5 must go between $4$ and $6$, so there are $3 \cdot 3! = 18$ ways.
$24 + 24 + 24 + 18 = 90$ ways if 4 is first.

If the first number is $3$, ....
3 6 _ _ _ _ $\implies$ 24 ways
3 _ 6 _ _ _ $\implies$ 24 ways
3 1 _ 6 _ _ $\implies$ 4 ways 
3 2 _ 6 _ _ $\implies$ 4 ways
3 4 _ 6 _ _ $\implies$ 6 ways
3 5 _ 6 _ _ $\implies$ 6 ways
3 5 _ _ 6 _ $\implies$ 6 ways
3 _ 5 _ 6 _ $\implies$ 6 ways
3 _ _ 5 6 _ $\implies$ 4 ways
$24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84$ ways

If the first number is $2$, ....
2 6 _ _ _ _ $\implies$ 24 ways
2 _ 6 _ _ _ $\implies$ 18 ways 
2 3 _ 6 _ _ $\implies$ 4 ways
2 4 _ 6 _ _ $\implies$ 6 ways
2 5 _ 6 _ _ $\implies$ 6 ways
2 5 _ _ 6 _ $\implies$ 6 ways
2 _ 5 _ 6 _ $\implies$ 4 ways
2 4 _ 5 6 _ $\implies$ 2 ways
2 3 4 5 6 1 $\implies$ 1 way

$24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71$ ways

Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$