discrete math: n pennies among k children with each child having atleast 2 pennies This problem is posed in Lovasz's Discrete math book chapter 3 and I understand the correct answer which is $$ \binom{n-k-1}{k-1} $$

However, why is my approach not right ? Here it goes.
We have n pennies. We group them into 2 each  and hence have 
$$
\lfloor{\frac{n}{2}}\rfloor
$$
units. Lets call them U. 
Now we have to distribute U units among k-1 children which will lead to

$$ \binom{\lfloor{\frac{n-2}{2}}\rfloor-2}{k-1} $$
ways. I subtracted 2 because 2 pennies will goto the first child.
Assume you have n pennies, and you want to distribute them among k children, with each child having at least 2 pennies. You can go as follow:

Distribute 2 pennies to k child.

Since the pennies are identical, the order you distribute the pennies to the children does not matter.

Then there is only one way to do that.

Now you left with n-2K pennies.

And the number of distributing the remaining pennies is:

$(n-2k)+k-1 \choose k-1 $ = $ n-k-1 \choose k-1$.
[By the previous theorem in the textbook].

So the total number is:

$ (1) $ $ n-k-1 \choose k-1$ = $ n-k-1 \choose k-1$.
You can see that your answer is wrong when different $n$ give the same answer.

What you should do is give to each child $2$ pennies.

You are left with $n-2k$ pennies and you want to give them to $k$ children.

Can you take it from here?

Recall that the number of ways to give $m$ pennies to $r$ children is
$$\binom{m+r-1}{r-1}.$$