In how many ways can you write a 6-digit number, with the digits in descending order, using digits from 1 to 6? Repetition of digits is allowed. Taking cases is far too lengthy. I also tried a binary approach, but that isn't helping my case either. 
 A: A number of the type we are trying to count is completely determined once we know the number of $6$'s, the number of $5$'s, and so on down to the number of $1$'s.
Let $x_1$ be the number of $1$'s, $x_2$ the number of $2$'s, and so on up to $x_6$. We want to find the number of solutions of the equation
$$x_1+x_2+\cdots+x_6=6$$
in non-negative integers.
Put in this way, the problem may already familiar, and you may know that the number of solutions is $\binom{6+6-1}{6-1}$ or equivalently $\binom{6+6-1}{6}$.
If this is not yet familiar, please see the article Stars and Bars in Wikipedia. There are also many references to Stars and Bars on MSE.
A: A six-digit number with the numbers written in descending order can be represented by placing six vertical bars in a row of six ones in such a way that the number of ones to the right of the bar represents the digit.  For instance,
$$| 1 || 1 | 1 1 | | 1 1$$
represents the number $655422$, while 
$$ 1 1 || 1 | 1 | 1 || 1 1$$
represents the number $443211$.  Since the last digit is at least $1$, the last symbol in the row must be a $1$.  Therefore, a particular choice of a six-digit number constructed from the digits $1, 2, 3, 4, 5, 6$ in which the digits appear in descending order is determined by choosing which six of the first eleven places in the row will be filled with bars, which can be done in $\binom{11}{6}$ ways. 
