Probability that the minimum is $2$ and the maximum is $5$ after rolling a die $4$ times 
Suppose that a fair $6$-sided die is rolled $4$ times. Letting $N$ be
  the minimum number of spots observed and $X$ be the maximum number of
  spots observed, give the value of $P(N= 2, X= 5)$

For this to happen, we need all $4$ die to be greater than or equal to $2$ but we need at least one to be $2$. Similarly, we need all $4$ die to be less than or equal to $5$ but we need at least one to be $5$. So I need to find the probability of getting a $2$, a $5$, and a $2,3,4$, or $5$, twice. I have 
$${4\choose{1}} \cdot{3\choose{1}} \cdot{2\choose{2}}\cdot {1\over{6}} \cdot{1\over{6}}\cdot {4\over{6}}^2 =.148$$
since we are choosing one spot of $4$ to be a $2$, one spot of the remaining $3$ to be a $5$, and $2$ spots of the remaining $2$ to be a $2,3,4,$ or $5$.
I do not think this is right because I did an excel simulation and got a probability of a little over $.09$.
 A: Your enumeration counts some rolls repeatedly.  For instance a 2 2 4 5 roll is counted twice, on account of the two different choices for 2.
I prefer an inclusion-exclusion-style argument for this, which yields a probability of 
$$
\frac{4^4-2*3^4+2^4}{6^4} = \frac{55}{648} \approx 0.0849
$$
Here the $4^4$ is all rolls with all values from 2 to 5, the two copies of $3^4$ are the rolls with all values from 2 to 4 and all rolls from 3 to 5, and the $2^4$ are the rolls with all values either 3 or 4.  You want to exclude values without a 5 and without a 2, but after subtracting the $2*3^4$ representing both of these possibilities, you need to add in the intersection of these two situations since those got subtracted twice.
A: I can see repetition in your argument. We have space for four rolls:
$$
----
$$
We choose one of these, and put a $2$ in there:
$$
-\ \color{red}2--
$$
We choose one of the remaining, and put a $5$ in there:
$$
-\ \color{red}2\  \color{blue}5\  -
$$
Put whatever of our choice in the remaining two:
$$
\color{green} 4 \ \color{red}2\  \color{blue}5\  \color{green}2
$$
However, there's way to get this in a different manner : start by putting a two in the fourth roll:
$$
---\ \color{red}2
$$
Proceed to put a $5$ in the same position as before:
$$
- - \color{blue}5\ \color{red}2
$$
Then fill the other two randomly:
$$
\color{green} 4 \ \color{green}2 \ \color{blue}5 \ \color{red} 2
$$
Two differently counted possibilities are the same. Therefore, your answer is bound to exceed the current answer.
Naturally, the way to correct this, is to remove such repetitions. You can do this by the inculsion-exclusion principle, as detailed in the above answer.
