Number of binary words that can be formed 
How many binary words of length $n$ are there with exactly $m$ 01 blocks?

I tried by finding number of ways to fill $n-2m$ gaps with $0$ and $1$ such that no $'01'$ block gets created again. But this method is not working and I am stuck in this problem. Please provide me an elegant solution of this problem.
Edit: Hw Chu has given a wonderful solution and I really appreciate this solution. But I am now interested in the intuition behind his solution. Further, I request to provide a relatively easier solution to this problem.

Here is another approach. It is "relatively easier" in the sense that setting up recurrence relations that condition on the last few digits of the binary word to force what we're interested in, is a standard way to approach such questions.        
Let $a_{n,m}$ be the number of binary words of length $n$ with exactly $m$ 01 blocks that end with a $0$.
Let $b_{n,m}$ be the number of binary words of length $n$ with exactly $m$ 01 blocks that end with a $1$.
We want to find $a_{n,m} + b_{n,m}$.   
Observe that the recurrance relations are:
$a_{n+1, m} = a_{n, m} + b_{n,m},$
$b_{n+1, m} = a_{n, m-1} + b_{n,m} $.
This follows directly because the only way to get an additional 01 block is to add a 1 to something in $a_{n, m}$.   
Claim: $a_{n,m} = { n \choose 2m+1}, b_{n,m} = { n \choose 2m}$.
(You can guess this by looking at several initial terms.)  
Proof: This can be shown by induction on $n$.
First, the base case of $n=1$ only involves $m=0$, and is clearly true.
For the induction step,
$a_{n+1,m} = a_{n,m} + b_{n,m} = { n \choose 2m+1} + { n \choose 2m} = {n+1 \choose 2m+1}$, and
$b_{n+1,m } = a_{n, m-1} + b_{n,m} = {n \choose 2m-1} + { n \choose 2m} = {n+1 \choose 2m}$.   
Hence, the number of binary words of length $n$ with exactly $m$ 01 blocks is  $a_{n,m} + b_{n,m} = { n+1 \choose 2m+1} $.   

You could modify this approach to count the number of   


*

*binary words with $m$ 11 blocks  (where 111 has 2 such blocks)    

*trenary words with $m$ 01 blocks   

*binary words with $m$ 010 blocks 

