how many words can be made with conditions on the number of times different letters are used (eg $n_1=2n_2$) I'm struggling with this one, but my latest attempt (which I've just realised is wrong is)
consider an alphabet ABCD, if A occurs twice as much as B then we have (for 4 letter results):
1 where A=B=0
3 where A=2 B=1 ($\frac{3!}{2!1!}1^1$)
I came to this by looking at "AABC, ABAC..." but the C here is fixed, at the end, it could go anywhere!
So I'm not sure how to do this (the question is for an alphabet of 4 with words of 7 were the frequency of the first is 2 times the frequency of the second)
could I simply multiply the "AABC" term by $\frac{4!}{2!}$ because that's how many times it can be re-arranged, this'd give 12*3 = 24 ways for 2As 1B (and 1C) which is actually precisely what I want...
If we introduce D it becomes more difficult though, 
So my actual question is:
How many ways are there to make words from a four letter alphabet of length 7 where the first letter occurs twice as much as the second
(that is the first and second occur 0 times, 2 and 1 respectively, or 4 and 2 respectively) The first one is just $2^4$ not sure where to go after that. 

Your  count for the $0$-$0$ case has a typo, it should be $2^7$. 
For $2$-$1$, we can choose where the $A$ goes in $\binom{7}{2}$ ways, and for each of these we can choose where the $B$ goes in $\binom{5}{1}$ ways, and then the rest in $2^4$ ways, so there are $\binom{7}{2}\binom{5}{1}2^4$ words of this type. 
For $4$-$2$ the same idea gives $\binom{7}{4}\binom{3}{2}2^1$. 
Add the counts for the three cases.
Remark: It is possible to give a general formula, but for a small number like $7$ the cases approach works nicely. 
