Probability of 3 of a kind with 7 dice Similar questions:
Chance of 7 of a kind with 10 dice
Probability of getting exactly $k$ of a kind in $n$ rolls of $m$-sided dice, where $k\leq n/2$
Probability was never my thing, so please bear with me.
I've reviewed the threads above to the best of my ability, but I still wonder how to go about finding a match of 3 from 7 dice.
At least three match, but no more (two sets of three is okay, a set of three and a set of four is not):
(a) : $ \frac{6  \binom{7}{3} 5^4}{6^7}  $ 
In the other discussions, this wasn't desired since it would allow for a second triple to occur, or even a quadruple. 
Odds of a quadruple with the remaining 4 dice:
(b) : $(1/5)^4 $
Then, the probability that from rolling 7 dice that there is at least three that match, and no more than three, would be:
(c) : $ \frac{6  \binom{7}{3} * 5^4}{6^7}- (1/5)^4 $
Exactly two sets of three:
(d) : $ \frac{6  \binom{7}{3}  \binom{4}{3} \binom{1}{1}}{ 6^7} $
Maybe?  My thought process was that if $\binom{7}{3}$ will give me a set of three, then with the remaining 4, I could pick 3 yielding $\binom{4}{3}$ with 1 leftover.  I realize this is probably wrong.  Why? What would be the proper way to go about this?
Exactly one set of three:
Then to find the probability that there is one and only one set of three from 7 dice, we could take the probability of one or more sets of three (c) and subtract the probability of exactly two sets (d), for:
$ \frac{6  \binom{7}{3}  5^4}{ 6^7} - (1/5)^4 - \frac{6  \binom{7}{3}  \binom{4}{3} \binom{1}{1} }{ 6^7} $
(e) : $  \left(\frac{6  \binom{7}{3}}{ 6^7}\right) \left( 5^4 -  \binom{4}{3} \binom{1}{1} \right) - (1/5)^4 $
Is this at all on the right path?
Thank you!
PS.
Sorry about the syntax, but I couldn't figure out how to make the standard nCr() symbol with MathJaX.

You are on the right track with $6^7 = 279936$ as the denominator. To find how many cases have three but no more matching, I would start by looking at the four partitions of 7 into up to 6 parts where the largest is 3: 3+3+1, 3+2+2, 3+2+1+1, 3+1+1+1+1.  You can then work out each systematically, taking account both the numbers that appear and the order they appear in.  
I think the first (which has two sets of three) is $$\frac{6!}{2!\;1!\;3!} \times \frac{7!}{3!\;3!\;1!} = 8400$$ which is ten times what you have in (c).  The others (just one three-of-a-kind) are 
$$\frac{6!}{1!\;2!\;3!} \times \frac{7!}{3!\;2!\;2!} + \frac{6!}{1!\;1!\;2!\;2!} \times \frac{7!}{3!\;2!\;1!\;1!} + \frac{6!}{1!\;4!\;1!} \times \frac{7!}{3!\;1!\;1!\;1!\;1!} = 113400$$. 
So I get about 0.405 for the probability of exactly one three-of-a-kind (but no four or more) and about 0.435 for the probability of the one or more threes-of-a-kind (but no four or more) 
