If 7 dice are thrown simultaneously, then what does the probability that all six digit appears on the upper face equal to? I've approached the problem the following way : 
Out of the 7 dice, I select any 6 which will have distinct numbers : 7C6.
In the 6 dice, there can be 6! ways in which distinct numbers appear.
And lastly, the last dice will have 6 possible ways in which it can show a number.
So the required answer should be : 7C6 * 6! * 6/(6^7) which on simplifying becomes : 70/(6^3 *  3).
However, the answer given is 35/(6^3 * 3).
Where exactly am I going wrong?

As noted in my comment above, you are double counting. Here's a different way of approaching the problem: let's compute the probability of throwing two 1's and one each of the other five numbers. So first choose the two dice that come up 1, in $\binom{7}{2}$ ways. The remaining five digits can be distributed in $5!$ ways, so the probability of throwing this combination is
$$\frac{5!\binom{7}{2}}{6^7}.$$
Thus the answer you are looking for is six times this, or
$$\frac{6!\binom{7}{2}}{6^7},$$
which comes out to $\frac{35}{6^3\cdot 3}$.
