Seven dice are thrown, what is the probability that all numbers show up on the dice?
My first answer uses the logic that if all numbers show up and you throw seven dice, then one number is repeated twice. Then there are $6\cdot\binom{7}{2}$ ways to choose the repeated number and then place them in all possible positions. For every unique position of these two repeated numbers there are $5!$ possible ways to obtain the other $5$ numbers on the dice. So using a counting argument, given that there are $6^{7}$ total permutations, the probability for the statement in the original question is:
$$\frac{6\cdot\binom{7}{2}\cdot5!}{6^{7}}$$
This is the answer I would go for, but I can't see the failure in my other argument which considers the first $6$ dice. Given that there is a unique number on every die, there are $6!$ possible permutations of this being possible. Now we add the seventh die, it can be a number from $1$ to $6$ and you can place it at any position in between the other six dice or at either end giving $7$ possible positions. This totals to $6!\cdot6\cdot7$ possibilities given a probability of:
$$\frac{6!\cdot6\cdot7}{6^{7}}$$
So my question, is the first argument correct, and if so, why is the second argument wrong?
The first argument is clear and correct.
For the second, consider for example the configuration $1234456$. You counted it twice. The "first six" could have beeb $123445$, and the "seventh" a $6$, left in place. Or the first six could have been $123446$, and the seventh a $5$, moved into the spot between $4$ and $6$.
A little more digging will show that every configuration is in fact double-counted. So the second method can be used, if it is followed by division by $2$.