Permutations on Array Elements (With Restrictions) This question is as follows:
"Let $K$ be an integer satisfying $1 \leq K \leq N$. In how many ways can we arrange the $N$ objects in $B$ so that $B_K$ is smaller than all the objects that follow it in $B$ (i.e., $B_K < B_I$ for all subscripts $I$ greater than $K$)? Answer the question for $N = 6$, $K = 4$ first. Then give the answer for general $N$ and $K$. Simplify your answer algebraically as much as possible."
My thought process and answer:
We have an array 6 elements long: [][][][][][]
(1) We need to place $3$ elements in the last positions so that the last $2$ are larger than the element in $4$th cell.  One element automatically goes in position $4$, so we don't need to calculate that, but (2) we need to arrange the elements in positions $5$ and $6$. (3) We now need to arrange the first $3$ elements.
(1) $6C3 \times$
(2) $2! \times$ 
(3) $3! = 240$
or
(1) $n C (n - k + 1) \times$
(2) $(n - k)! \times$
(3) $(n - k + 1)!$    
My formula is derived from the items in the parenthesis above:
$$\frac{n!(n-k)!(n - k + 1)!}{(n-k+1)!\bigl(n-(n-k+1)\bigr)!}$$
and I reduced it down to:
$$\frac{n!(n-k)!}{(k-1)!}$$
Although, I can't tell if it can be reduced anymore.  I also am not confident at all in my numerical answer.
 A: Your answer for the case $N = 6$, $K = 4$ is correct.
Your answer for the general case is not.  Consider $n = 7$ and $k = 2$.  Your formula 
$$\frac{n!(n - k)!}{(k - 1)!}$$
yields
$$\frac{7!5!}{1!} = 7!5! > 7!$$
which is impossible there are $7!$ ways to arrange $7$ objects without restrictions.  The reason is that your factor of $(n - k + 1)!$ is incorrect since only $k - 1$ objects appear before the $k$th object.
We can choose which $k - 1$ objects appear before the $k$th position in $\binom{n}{k - 1}$ ways, permute them in $(k - 1)!$ ways, and permute the $n - k$ objects that appear after the $k$th position in $(n - k)!$ ways.  Hence, the number of permissible arrangements 
\begin{align*}
\binom{n}{k - 1}(k - 1)!(n - k)! & = \frac{n!}{(k - 1)![n - (k - 1)]!}(k - 1)!(n - k)!\\ 
& = \frac{n!}{(k - 1)!(n - k + 1)!}(k - 1)!(n - k)!\\ 
& = \frac{n!}{n - k + 1}
\end{align*}
Alternatively, we can select which $n - (k - 1) = n - k + 1$ objects which appear in the last $n - (k - 1)$ positions in $\binom{n}{n - k + 1}$ ways, permute the $n - k$ objects after the $k$th position in $(n - k)!$ ways, and permute the first $k - 1$ objects in $(k - 1)!$ ways.  Hence, the number of permissible arrangements is 
$$\binom{n}{n - k + 1}(n - k)!(k - 1)! = \frac{n!}{(n - k + 1)!(k - 1)!}(n - k)!(k - 1)! = \frac{n!}{n - k + 1}$$
