Is there a simpler way to calculate this? 
A number of five different digits is written down at random. Find the probability that the number has the largest digit in the middle.

My Solution:
If $9$ is in the middle then the first place from left can be filled in $8$ ways. Remaining $3$ places can be filled in $^8P_3$ ways.
If $8$ is in the middle then the number of ways are $7\times (^7P_3)$
This would go till $3(^3P_3)$
Thus, probability$=\frac{\sum_{n=3}^8n(^nP_3)}{9(^9P_3)}$
I calculated using calculator and got $\frac7{36}$
Is there an easier way to calculate this (in a time bound exam)?
I tried writing $n(^nP_3)$ as $(n+1-1)(^nP_3)=(n+1)(^nP_3)-(^nP_3)=(n+1)(n)(n-1)(n-2)-(n)(n-1)(n-2)$
I was hoping to get telescopic series, but not getting that.
Another try:
$n(^nP_3)=\frac{n\times n!}{(n-3)!}\times\frac{3!}{3!}=6n(^nC_3)=6(n+1-1)(^nC_3)=6(4(^{n+1}C_4)-(^nC_3))$
Not able to conclude from this.
 A: If you allow a leading zero then there are $\frac{10!}{5!}=30240$ possible numbers with distinct digits, of which a fifth, i.e. $6048$, have the largest in the third place
Of these, a tenth have a leading zero, i.e $3024$, of which a quarter, i.e. $756$, have the largest in the third place
So prohibiting a leading zero, that leaves $27216$ possible numbers with distinct digits, of which $5292$ have the largest in the third place.  $\frac{5292}{27216}=\frac{7}{36} \approx 0.194$
You can do this with smaller fractions rather than counts: $$\dfrac{\frac15  - \frac14 \times \frac1{10}}{1-\frac1{10}} = \dfrac{8-1}{40-4}=\dfrac{7}{36}$$
A: If you allow leading zeroes, then this is simply $\frac{1}{5}$ as every digit is equally likely to appear at each position.
If you do not allow leading zeroes, then the probability gets a bit worse as if a zero digit appears then it is more likely to be the center digit than other digits due to the restriction of zero not being allowed as the leading digit.
If you do not allow leading zeroes, then I'd break into cases based on if a zero was or was not used.  For those where zero is not used, again the simple answer of $\frac{1}{5}$ applies.  Using your notation, no zero appears with probability $\frac{~^9P_5}{9\cdot ~^9P_4}$
For those where zero is used, first check that the zero doesn't go in the middle which occurs $\frac{3}{4}$ of the time, and then given that happened check that of those nonzero terms the largest goes into middle which occurs $\frac{1}{4}$ of the time, for a combined probability of $\frac{3}{16}$.  The probability of a zero appearing is $\frac{4\cdot ~^9P_4}{9\cdot ~^9P_5}$
This gives a final answer of:
$$\frac{~^9P_5}{9\cdot ~^9P_4}\cdot \frac{1}{5} + \frac{4\cdot ~^9P_4}{9\cdot ~^9P_4}\cdot \frac{3}{16} = \frac{7}{36}$$
