Total number of combinations for boxes that can hold a maximum of 2 objects. Consider an $N$ balls than can be placed in $n$ number of boxes. 
Where: $n > N/2$
Suppose each boxes can only hold a maximum of two balls, what is the formula for finding the total number of combinations if each balls are considered distinguishable?
 A: Since the boxes are indistinguishable we have to count the number of ways we can pair off some of $n\geq1$ distinguishable people $P_k$, $1\leq k\leq n$. Denote this number by $a_n$. Then $a_1=1$, $a_2=2$, and we have the following recursion:
$$a_{n+1}=a_n+ n\>a_{n-1}\ .$$
The reason is as follows: You can decide to keep person $P_{n+1}$ alone and pair off the remaining $n$ persons at leisure, or you can pair $P_{n+1}$ off with a chosen person $P_k$, $1\leq k\leq n$, and then pair off the remaining $n-1$ persons at leisure.
The resulting sequence is
$$1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496, 35696, 140152, 568504, \ldots\ ,$$
which is A000085 at OEIS.
A: Your question can be reformulated: Find the number of partitions of a N-set into n parts each containing at most two elements.$$\sum_{i=N-n}^{N/2}\frac{N!}{i!2^i(N-2i)!}$$
Example 1:
For $N=4,n=2$ from last formula we get $$\frac{4!}{2!2^20!}=3$$ the partitions are
$$(ab|cd),(ac|bd),(ad|bc)$$
Example 2:
For $N=3,n=2$ from last formula we get $$\frac{3!}{1!2^11!}=3$$ the partitions are
$$(ab|c),(ac|b),(a|bc)$$
Example 3:
For $N=4,n=3$ from last formula we get $$\frac{4!}{1!2^12!}=6$$ the partitions are
$$(\emptyset|ab|cd),(\emptyset|ac,bd),(\emptyset|ad|bc)$$
$$(a|b|cd),(a|c|bd),(a|d|bc)$$
