chances of getting three of one kind and four of another out of seven dice There are several questions similar to this one but after reading those, 
I am still very confused.
I also did a similar problem of this one and I think I got it, but then I got stuck again.
So if four dice are rolled, the chance of getting three of a kind is:
$ \binom{6}{1} \frac{1}{6}* \binom{5}{1} \frac{1}{6}*\binom{4}{1} \frac{1}{6} *\frac{1}{6}$
so if seven dice are rolled, in my understanding, 
the chance of getting three of a kind and four of another would be:
$ \binom{7}{1} \frac{1}{6}*\binom{6}{1} \frac{1}{6}*\binom{5}{1} \frac{1}{6} *\binom{4}{1} \frac{1}{6}  *\binom{3}{1} \frac{1}{6} *\binom{2}{1} \frac{1}{6} *\binom{1}{1} \frac{1}{6}  $
however, the answer in the book is $ \frac{6  *5*\binom{7}{4} }{6^7} $ and I am totally lost.
Please help!
additional problems
The answers you guys gave kind of make sense to me but they also make me very confused.
can I think of it using the way I did above? 
for example, another part of the questions asked about the chance of getting two fours, two fives and three sixes.
I think of it as:
$ \binom{7}{2} \frac{1}{6}^2*\binom{5}{2} \frac{1}{6}^2*\binom{3}{3} \frac{1}{6}^3  $ which matches the solution in the book.
 A: There are $6^7$ possible outcomes, listed as 7-tuples. Now, how many 7-tuples would consist of two distinct elements, 4 of one kind, and 3 of other. Such 7-tuples can be ordered into $[a,a,a,a,b,b,b]$ with $a \not= b$. $a$ can be chosen 6 different ways. Once $a$ is chosen, $b$ can be chosen 5 different ways. There are $\binom{7}{4}$ ways to rearrange $[a,a,a,a,b,b,b]$ tuple. 
The probability is then the ratio of the number of needed outcomes, over the number of total outcomes:
$$
  p = \frac{1}{6^7} \binom{7}{4} 6 \times 5
$$
A: Think of filling in 7 slots; in each you have the value of a die roll.
There are $6\cdot 5$ ways to choose the values for the three and, different valued, four of a kind (for example,
the three of a kind is three '2's and the four of a kind is four '5's).
There are $7\choose 4$ ways to select ''slots'' in which to place your 4 of a kind.
The remaining three slots will then contain the three of a kind. So, there are
$6\cdot5\cdot{7\choose 4}$ different ways to obtain a  three of a kind and 4 of a kind.  Since outcomes are equally likely here, the chance of obtaining  a  three of a kind and 4 of a kind is as you stated.
