Racko!

Obviously the 60 cards do not matter, curiously the fact that 10 cards are dealt doesn't (for this problem but I'll call it $C$ below) either.

Let $P(x)$ be the probability of $x$ cards in ascending order.

$P(0)=0$ because some card has to be first

$P(1)=\frac{1}{2}$ because there's an equal chance that the first card is smaller than the second.

From here we have the general formula

$\large P(x)= \begin{cases} \frac{x}{(x+1)!}, & \mbox{if } x<C \\ \frac{1}{(x)!}, & \mbox{if } x=C \end{cases}$

Proof: Suppose you were given cards and asked for the probability that the first $(x+1)$ were in order. That would be $\frac{1}{(x+1)!}$ . Now, for each of those, take any one of the first $x$ cards and move it to the last position. Those are all of the possibilities for first $x$ in order and next one lower given some set of $x+1$ cards to start. Thus, $P(x)=\frac{x}{(x+1)!}$ unless there is no last card because $x=C$ , but then $P(x)=\frac{1}{(x)!}$ is obvious.

In the problem above $x=6$ so $P(6)=\frac{6}{(7)!} = \frac{1}{840}$ so $\boxed{n=840}$