Find the number of ways of cooking dishes subject to these restrictions I have the following problem: In $N$ days, I have to do at most $K$ tasks given that I can do only single task in one day and can do every task in $M$ ways. In how many ways can I do this? For instance, I have $N=1$, $K=1$ and $M=1$.
The problem can be interpreted as that in $N=1$ day, I have to do at most $K=1$ task given that I can do that task in $M=1$ ways. So what are the number of ways I can do this task and the answer for this is $1$. But I am confused over the general formula for the problem.
For $N=4$, $M=3$, $K=2$, the answer comes out to be $45$, but I am get baffles on seeing the answers.
Please help me out. Thanks in advance.

Added later. This is the exact question; can you now help me about it?

Chef Dengklek will open a new restaurant in the city. The restaurant will be open for $N$ days. He can cook $M$ different types of dish. He would like to cook a single dish every day, such that for the entire $N$ days, he only cook at most $K$ distinct types of dishes. In how many ways can he do that?

 A: Answer edited in light of comment: In your Chef Dengklek question, if $M=K$, then the number of possible ways of cooking up to $K$ different dishes in $N$ days would be $K^N$.  So the number of ways of cooking exactly $K$ different dishes would be $K^N-{K \choose 1}(K-1)^N+{K \choose 2}(K-2)^N-\cdots$; this can also be written as $K! S_2(N,K)$ where $S_2$ are Stirling numbers of the second kind.
So allowing $M\ge K$,  the number of ways of cooking exactly $K$ different dishes would be ${M \choose K}K! S_2(N,K)$ so the total number of possible ways of cooking up to K different dishes in N days is $$\sum_{i=1}^K {M \choose i}i! S_2(N,i).$$    
For $N=4$, $M=3$, $K=2$, this gives $${3 \choose 1}1!\times S_2(4,1)+ {3 \choose 2}2!\times S_2(4,2) = 3\times 1 \times 1 + 3\times 2 \times 7  = 3+42=45.$$
A: I, too, am baffled by 45. First, choose the 2 days when you'll do the tasks; this can be done in $4\times3=12$ ways. 3 ways to do each task means multiply by a couple of threes, getting $12\times3\times3=108$. And this is assuming you do both tasks - you say, "do at most $K$ tasks," so you have to add in all the ways of just doing one task, and the one way of not doing any task. So I get a lot more than 45.  
A: I will assume that you have to do exactly $K$ tasks.Under this assumption general formula is:
$$n=\frac{N!}{(N-K)!}\cdot M^{K}$$
where $n$ is number of  ways.
As Gerry pointed out you have to add in all those solutions when number of tasks is less than $K$ in order to get number of all possible ways.
EDIT:
Number of all possible ways is given by :
$$ \sum_{i=0}^K \frac{N!}{(N-i)!}\cdot M^{i}$$ 
