Principle of inclusion-exclusion when counting number of unique faces We throw a die a number of times. We let $Y_k$ be the number of different faces that came up after $k$ throws. Now we want $P[Y_k \leq 5]$.
According to the answer sheet, we use the principle of inclusion-exclusion here to arrive at the following answer:
$P(Y_k \leq 5) = \binom{6}{5}(\frac{5}{6})^k - \binom{6}{4}(\frac{4}{6})^k +  \binom{6}{3}(\frac{3}{6})^k - \binom{6}{2}(\frac{2}{6})^k + \binom{6}{1}(\frac{1}{6})^k - \binom{6}{0}(\frac{0}{6})^k$
(which I believe to be equal to)
$P(Y_k \leq 5) = P(Y_k = 5) - P(Y_k = 4) + P(Y_k = 3) - P(Y_k = 2) + P(Y_k = 1) - P(Y_k = 0)$
But I just can't wrap my head around how the principle of inclusion-exclusion was exactly used here. My first thought was that it has something to do with the fact that if we see 5 faces, we've also seen at least 4 faces, etc. so that $[Y_k \leq 4]$ is a subset of $[Y_k \leq 5]$.  I've tried to manually calculate (using the principle of inclusion-exclusion) $P(Y_k = 1 \cup Y_k = 2 \cup Y_k = 3 \cup Y_k = 4 \cup Y_k = 5)$ but I don't arrive at the same answer.
So far I haven't come to a good intuitive understanding yet, it would be great if someone could help me grasp this.
 A: Let $E_i$ be the event that face number $i$ does not appear, for each $i=1,2,\dots,6$. Then event $\{Y_k\le 5\}$ is the same as the event that at least one $E_i$ occurs, i.e. that at least one face is missing. Therefore, using the principle of inclusion exclusion, and the symmetry of the problem,
\begin{align}
P(Y_k\le 5)
&=P\left(\bigcup_{i=1}^6E_i\right)
\\&=\sum_{r=1}^6(-1)^{r+1}\binom{6}rP(E_1\cap E_2\cap\dots\cap E_r)
\\&=\sum_{r=1}^6(-1)^{r+1}\binom6r\left(\frac{6-r}{6}\right)^k.
\end{align}
This is where the equality you were given comes from.
A: No, you are misinterpreting.  The equation you have is
$$
\mathbb{P}(\lvert F\rvert\leq 5)=
\sum_{\lvert A\rvert=5}\mathbb{P}(F\subseteq A)
-\sum_{\lvert A\rvert=4}\mathbb{P}(F\subseteq A)
+\sum_{\lvert A\rvert=3}\mathbb{P}(F\subseteq A)
-\sum_{\lvert A\rvert=2}\mathbb{P}(F\subseteq A)
+\sum_{\lvert A\rvert=1}\mathbb{P}(F\subseteq A)
-\sum_{\lvert A\rvert=0}\mathbb{P}(F\subseteq A)
\tag{1}
$$
where $F$ is the set of faces that show up, and $A\subseteq \{1,2,3,4,5,6\}$ is a subset of faces in each sum.

Expanding a little more, we have 6 events "only 1,2,3,4,5 are allowed", "only 1,2,3,4,6 are allowed", ..., "only 2,3,4,5,6 are allowed", or equivalently, the six events are "1 does not show up", "2 does not show up", ..., "6 does not show up".  $\lvert F\rvert\leq 5$ is the union of these events, and intersecting $k$ of these events is precisely reducing the allowable faces to 6-$k$.  So inclusion-exclusion gives
$$
\mathbb{P}(\lvert F\rvert\leq 5)=\sum_{\text{possible }1\text{-intersection}}\mathbb{P}(event) - \sum_{\text{possible }2\text{-intersection}}\mathbb{P}(event)+\dots
$$
which is equation (1).
A: The first term, ${6\choose5}\left(\frac56\right)^k$, represents the probability that only five faces came up.  There are ${6\choose5}$ ways to choose which of the faces come up.  The problem is that we have double-counted the cases where only four faces came up.  For example, if only $1,2,3,4$ show up, we count this both as only $1,2,3,4,5$ showing up and as only $1,2,3,4,6$ showing up.  So we have to subtract the cases where only four faces show up.  This accounts for the second term.  But then there is a double-counting problem when only three faces show up, and so on.
This is a typical application of the principal of inclusion and exclusion.
EDIT
When I say "five faces come up" I'm being a little sloppy.  What I mean is that all the faces that show up belong to some set of five faces.  This doesn't exclude the possibility that only four or fewer faces actually show up.  So as you as in your comment, if the faces that show up belong to $\{1,2,3,4\}$ we've counted that once in the $\{1,2,3,4,5\}$ case and one in the $\{1,2,3,4,6\}$ case.  
Now if you analyze the $\{1,2,3\}$ case, you see that we've counted it three times and subtracted it three times, so we have to add it back in.
A: I usually use a counting method to solve problems like this so I'll use the formula posted by Mike Ernest to solve the following specific problem. A die is rolled $10$ times, what is the probability that at most only five numbers from six show up?
$P =\sum_{r=1}^6(-1)^{r+1}\binom6r\left(\frac{6-r}{6}\right)^{10}$
$= 6\cdot (\frac{5}{6})^{10} - 15\cdot (\frac{4}{6})^{10} + 20\cdot (\frac{3}{6})^{10} - 15\cdot (\frac{2}{6})^{10} + 6\cdot (\frac{1}{6})^{10} - 1\cdot (\frac{0}{6})^{10}$
$= .969033 - .260123 + .019531 - .000254 + 0 - 0\ $ (last two terms too small to count)
$P = .7282$
