Rencontres Number (Counting partial derangements)

Given two numbers, n >= 0 and 0 <= k <= n, count the number of derangements with k fixed points.
Examples: 
 

Input : n = 3, k = 0
Output : 2
Since k = 0, no point needs to be on its
original position. So derangements
are {3, 1, 2} and {2, 3, 1}

Input : n = 3, k = 1
Output : 3
Since k = 1, one point needs to be on its
original position. So partial derangements
are {1, 3, 2}, {3, 2, 1} and {2, 1, 3}

Input : n = 7, k = 2
Output : 924

In combinatorial mathematics, the rencontres number< or D(n, k) represents count of partial derangements.
The recurrence relation to find Rencontres Number D_{n, k}:
 

D_{(0, 0)} = 1 
D_{(0, 1)} = 0 
D_{(n+2, 0)} = (n+1) * (D_{(n+1, 0)} + D_{(n, 0)}) 
D_{(n, k)} = ⁿC_k * D_{(n-k, 0)})

Given the two positive integer n and k. The task is find rencontres number D(n, k) for giver n and k.
Below is Recursive solution of this approach:
 

924

Time Complexity: O(n * m), where n and m represents the given integers.
Auxiliary Space: O(n*m), due to recursive stack space.

Below is the implementation using Dynamic Programming: 
 

924

Time Complexity: O(n * m), where n and m represents the given integers.
Auxiliary Space: O(n * m), where n and m represents the given integers.