Probability of guessing $0 \le k \le 6$ numbers and their positions correctly, if 6 numbers are drawn from a pool of 49 numbers. Suppose 6 numbers are drawn one after another from a pool of 49 distinct  numbers. Once a number has been drawn, it is not put back in the pool. The number itself and its position are noted. You are now asked to guess 6 distinct numbers from 1 to 49.
What is the probability that $0 \le k \le 6$ of the numbers you guessed are the correct number and in the correct position and that the other $6-k$ numbers are either in the wrong position, or not among the drawn numbers?
I'm not sure what the combinatorial argument is, to calculate this probability correctly. 
There are $49^\underline{6} = 49*48*\dotsc*44$ possible ways to choose 6 numbers and hence the probability of guessing all six numbers correctly (in the correct order) is $1/49^\underline{6}$.
However, what happens, for example, if $k=4$? Do I have to simply multiply this probability by the number of ways to select 4 out of 6 numbers and the number of ways to choose 2 numbers from the remaining 43 numbers or do I still have to take the ordering of these 4 or 2 numbers into account?
I'm not sure which of the following two terms is correct or if they are both equally wrong (for $k = 4$):
\begin{align*}
\frac{{6 \choose 4}{43 \choose 2}}{49^\underline{6}}, \quad \frac{{6 \choose 4}4!{43 \choose 2}2!}{49^\underline{6}}
\end{align*}
For clarification: Let 1,2,3,4,5,6 be the numbers drawn from the pool in this order.


*

*If $k=6$ then the only valid guess is 1,2,3,4,5,6 (2,1,3,4,5,6 would be not)

*If $k=3$ then 1,2,3,6,4,5, 1,2,3,43,20,10, 43,2,3,4,20,10, 43,20,10,4,5,6 are all valid but a guess like 1,2,4,43,20,10 is not (but would be for $k=2$).

*If $k=0$ then all numbers from 1 to 6 must be at different positions or not included in the guess.

 A: As was already pointed out in the comments, both suggested formulas are wrong. To obtain the general expression let us proceed as follows.
Consider an auxiliary problem. Let $m$ bins be numbered from $1$ to $m$ and $n$ balls be numbered from $1$ to $n$ ($n\ge m$). In how many ways can we place one ball per bin so that no bin contains the ball with the number of the bin?
Obviously if there were no restriction there are $\frac{n!}{(n-m)!}$ ways to place one ball per bin. From this number we should subtract the number of combinations containing ball $\#X$ in the bin $\#X$ for each $X$ in the range $1..m$. In this way we will however repeatedly subtract the combinations containing both the ball $\#X$ in the bin $\#X$ and the ball $\#Y$ in the bin $\#Y$, which we should now add to the previous result. Proceeding in this way (keyword: inclusion-exclusion principle) one obtains:
$$
D(n,m)=\sum_{i=0}^m (-1)^{m-i}\binom mi\frac{(n-i)!}{(n-m)!}.\tag1
$$
Now we can proceed the original problem. To count the number of $m$-tuples with exactly $k$ coincidences (both of value and position) with a given one let us place $k$ balls (numbered in the range $1..m$) in the bins with the corresponding number. The rest $m-k$ bins we shall fill according to the restriction of the above auxiliary problem with balls chosen from the rest $n-k$ ones. Therefore the number in question is:
$$
N(n,m,k)=\binom mk D(n-k,m-k)=\binom mk\sum_{i=0}^{m-k} (-1)^{m-k-i}\binom {m-k}i\frac{(n-k-i)!}{(n-m)!}.\tag2
$$
In your case one obtains:
$$\begin{align}
N(49,6,0)&=8897907877,\\
N(49,6,1)&=1109783394,\\
N(49,6,2)&=58922175,\\
N(49,6,3)&=1705420,\\
N(49,6,4)&=28395,\\
N(49,6,5)&=258,\\
N(49,6,6)&=1.
\end{align}
$$
