Permutations of a sequence "Given a set $\{1,\ 2,\ 3,\ 4\}$, how many sequences with a length of $4$ with entries from this set have exactly one entry equal to $1$?"
Here is my work so far:
$$X = \left\{\text{sequences with length 4 from}\ \{1, 2, 3, 4\}\ \text{with exactly one entry equal to $1$}\right\}$$ $$Y = \left\{\text{permutations of length 4 from}\ \{1, 2, 3, 4\}\right\}$$  
By definition, $|Y| = 4\times 4\times 4\times4 = 256$.
Where do I go from here? Should I find $|X|$ as well? I think (by intuition, I haven't checked) that I need to find $|X \cap Y|$ in order to solve this problem. Is this true?
 A: The $1$ can be in any one of $4$ places. For each placement of the $1$, there are $3^3$ ways to fill in the rest of the entries. 
Note that the problem did not say that the other entries are distinct. It said only that they are not $1$. So for example $4143$ is one of our sequences.
A: Let’s straighten out your terminology first: permutations of $\{1,2,3,4\}$ are automatically of length $4$, and there are $4!$ of them, not $4^4$; there are $4^4$ $4$-term sequences of elements of the set $\{1,2,3,4\}$.
You want nothing to do with permutations of $\{1,2,3,4\}$, and you certainly need to find $|X|$: that’s what the question asks for! To do this, note that each $4$-term sequence in $X$ can be constructed by deciding first which term of the sequence is to be the $1$ and then what members of $\{2,3,4\}$ are to be assigned to the other terms of the sequence. There are $4$ ways to choose the term that’s to be $1$. Once that’s settled, each of the other terms can be given any of $3$ values, $2,3$, or $4$, so it takes $3$ $3$-way choices to pin down the rest of the sequence. Thus, we end up with $4\cdot3\cdot3\cdot3=108$ possible sequences.
