Question: A bag contains $n$ balls numbered $1$ to $n$.

We draw ALL the balls one by one till we empty the bag.

**Question 1** : What is the probability that the balls numbered $1$ , $2$, $3$ get drawn consecutively and in this order.

*My answer :*
$$\dfrac{(n-3)!}{n!}$$

**Question 2** : What is the probability that the balls numbered $1$ , $2$, $3$ get drawn in this order but not necessarily consecutively.

*My answer :*
$$\dfrac{\binom n3 \times (n-3)!}{n!}$$

explanation (for both) : 

$\binom n3$ number of ways to choose three positions for our three balls in $n$ locations

$n!$ total number of ways to draw all balls from the bag.

$(n-3)!$ number of ways to order the remaining balls between one another.

Is my analysis sound and correct ?

thanks.

Answer: 
I think you're almost there. For the first question, you might consider balls 1, 2, 3 as one "big ball". Then you would have $n-2$ total balls. There are $(n-2)!$ ways to arrange these balls and precisely 1 way to rearrange the balls in the "big ball". You can get the probability from here.

For the second exercise, you seem to have the right answer and justification. My reasoning would be that we have $n \choose 3$ ways to choose the positions of 1, 2, and 3. There is only one way to arrange these balls. Then there are $(n-3)!$ ways to rearrange the rest of the balls, giving the answer you provided.
**Question 1:**

You're answer for this question is almost correct. The only problem with your answer is that it doesn't take into account that there are $n-2$ places where you could start to draw your consecutive $1,2,3$. Therefore the answer should be:
$$\dfrac{(n-2)!}{n!}$$

**Question 2:**

The answer you arrived at(which simplifies to $\frac{1}{3!}$) is correct, but personally I feel like there is a slightly easier way to think about this problem. 

Since we are arbitrarily drawing our balls from $1$ to $n$, each permutation of $1$,$2$,$3$ is equally likely to be drawn therefore the probability that we draw them in the order $1$,$2$,$3$ is $\frac{1}{3!}$.