Probability Help (die problem) A die is rolled 20 times. How many different sequences 
a) each number 1-6 is rolled exactly three times
My Answer: (20 choose 6)*(3 choose 1)
b) each number 1-6 are each rolled exactly once in the first six rolls?
My Answer: (20 choose 6)*(6 choose 1)
c) each number rolled is at least as big as the number that was rolled directly before it?
ex: 111222333344455566666
My Answer: No idea
 A: Each number 1-6 is rolled exactly three times
No sequence of twenty rolls can be created out of 18 numbers.

Each number 1-6 are each rolled exactly once in the first six rolls
Consider just the first six rolls.  The first roll can be anything.  The second roll can be any of the five remaining numbers, the third roll can be any of the four remaining numbers, etc.  There are $6!$ such sequences.
The remaining fourteen rolls can be any of the $6^{14}$ sequences.
Therefore, there are $6!\ 6^{14}$ sequences of twenty rolls that fit the distinctness requirement for the first six rolls.

Each number rolled is at least as big as the number that was rolled directly before it
Let $M(\mathscr{l},s)$ be the number of sequences of length $\mathscr{l}$ with $s$ monotonically progressing symbols.  We want to find $M(20,6)$.  Either the first element of the sequence is ⚅ (six) or it isn't ⚅ (six), so we have:
$$ M(20, 6)\ =\ M(19, 6)\ +\ M(20, 5) $$
where $M(19, 6)$ is the number of sequences that start with ⚅ (six), and $M(20, 5)$ is the number of sequences that start with ⚄ (five) or smaller.
In general, we can set up a recurrence relation:
$$ M(\mathscr{l}, s)\ =\ M(\mathscr{l}-1, s)\ +\ M(\mathscr{l}, s-1)$$
with base cases
$$\begin{eqnarray*}
M(\mathscr{l}, 1) &= 1 \\
M(1, s) &= s \\
\end{eqnarray*}$$
Graphically illustrated, the problem is:
$$
\newcommand{r}[0]{\rightarrow}\newcommand{d}[0]{\downarrow}
\begin{array}{c}
M(20,6) &\r& M(20,5) &\r& \cdots &\r& M(20,1)=1 \\
  \d    &  &   \d    &  &        &  &   \d      \\
M(19,6) &\r& M(19,5) &\r& \cdots &\r& M(19,1)=1 \\
  \d    &  &   \d    &  &        &  &   \d      \\
M(18,6) &\r& M(18,5) &\r& \cdots &\r& M(18,1)=1 \\
  \d    &  &   \d    &  &        &  &   \d      \\
\vdots  &  & \vdots  &  & \ddots &  & \vdots    \\
M(1,6)=6&\r&M(1,5)=5 &\r& \cdots &\r&  M(1,1)=1 \\
\end{array}
$$
That's the same as saying that $M(\mathscr{l},s)$ is the number of paths, moving only downward and rightward, from $(\mathscr{l},s)$ to $(0,1)$, if we artificially extend the grid by one row.
$$
\newcommand{r}[0]{\rightarrow}\newcommand{d}[0]{\downarrow}
\begin{array}{c}
M(20,6) &\r& M(20,5) &\r& \cdots &\r& M(20,1)=1 \\
  \d    &  &   \d    &  &        &  &   \d      \\
M(19,6) &\r& M(19,5) &\r& \cdots &\r& M(19,1)=1 \\
  \d    &  &   \d    &  &        &  &   \d      \\
M(18,6) &\r& M(18,5) &\r& \cdots &\r& M(18,1)=1 \\
  \d    &  &   \d    &  &        &  &   \d      \\
\vdots  &  & \vdots  &  & \ddots &  & \vdots    \\
M(1,6)=6&\r&M(1,5)=5 &\r& \cdots &\r&  M(1,1)=1 \\
  \d    &  &   \d    &  &        &  &   \d      \\
M(0,6)=1&\r&M(0,5)=1 &\r& \cdots &\r&  M(0,1)=1 \\
\end{array}
$$
The answer to that combinatoric problem is $\left(\begin{array}{c}W+H\\W\end{array}\right) = \left(\begin{array}{c}W+H\\H\end{array}\right)$, where $W$ is the width (the number of right arrows) and $H$ is the height (the number of down arrows in the extended graph).  To convince yourself: every path from the top-left to bottom-right corner is a sequence of $W + H$ steps, of which $W$ have to be rightward steps.
Therefore, in general,
$$M(\mathscr{l},s) = \left(\begin{array}{c}\mathscr{l}+s-1\\s-1\end{array}\right) = \left(\begin{array}{c}\mathscr{l}+s-1\\ \mathscr{l}\end{array}\right)$$
and $M(20,6) = \left(\begin{array}{c}25\\5\end{array}\right) = 53130$.
A: a) I am considering it to mean atleast 3 times, which can be obtained by an exponential generating function:
\begin{align*}
  G(x) &= \left(\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\right)^6 \\
  \left[\frac{x^{20}}{20!}\right]G(x) &= \frac{11}{414720}\cdot 20! = 64530097632000
\end{align*}
b) has been answered : $$6!\, 6^{14} = 56422198149120$$
c) Use a program to generate the sequence : $$6,21,56,126,252,462,792,1287,2002,\ldots$$
Looking up in OEIS gives A000389.
Hence, for $n$ dice throws, the number of valid sequences are:
\begin{align*}
  a_n &= \binom{n+5}{5} \\
  \implies a_{20} &= \binom{25}{5} = 53130
\end{align*}
A: There is sort of a solution to c. It is preposterously long and cumbersome, maybe it will help to find something shorter.
$\sum _{a=1}^z \sum _{b=a}^z \sum _{c=b}^z \sum _{d=c}^z ...\sum _{q=p}^z \sum _{r=q}^z \sum _{s=r}^z \sum _{t=s}^z 1$
when z = 6 we get 53130.
