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?

You probably noticed that your answer differs from the correct answer by a factor 2, so apparently you count everything twice.
Suppose your dice are labeled A, B, C, D, E, F, G and you throw:
A:1
B:2
C: 3
D:4
E: 5
F: 6
G: 1
Then you count this throw twice: one time with ABCDEF as the 'special' dice showing 6 different figures and G as the redundant die, and once with BCDEFG as the special dice showing 6 different figures and A as the redundant die.
