How many possible combinations in 8 character String? Assume that we have $52$ characters; $26+26$ lowercase and uppercase letters.
We want to create passwords of eight characters long.
There are two conditions


*

*The letter I (uppercase i) has to be included.

*Characters may appear only once.


So, for instance "aaaaaaIaa" is illegal, however "aIbcdefgh" is OK. 
My solution: 
Since no character may be  duplicated and  the string must contain at least one  I , we have the following basic forms:
I******** 
*I*******
**I******
***I*****
****I****
*****I***
******I**
********I
Then the possible  combination  would be  $ 8*Perm_{7}(51) = 8* \frac{51!}{(51-7)!} $ 
Is that correct ? 
 A: Include 'I', select $7$ characters and them permute the $8$ letters. So,
Number of possible passwords
$=\dbinom{51}{7}\times 8!$ 

Proof:
Take a small example. Suppose we have the letters (a,b,c,d) to select from and we are counting the possible arrangements of $3$ letters where c must be come in each arrangement.
According to the same principle used, 
$\dbinom{3}{2} \times 3!=18$ arrangements are possible. These are

abc,acb,bac,bca,cab,cba  acd,adc,cad,cda,dac,dca
  bcd,bdc,cbd,cdb,dbc,dcb

Accoding to some other answers, it must be $\dbinom{3}{2} \times 3=9$ which is clearly not the right way of approaching this problem.
A: Your solution is correct. 
Multiplying with $8$ is OK, and the "basic forms" are OK. The question remains: How many possibilities are there to place the $51$ remaining characters in to the $7$ remaining places.
The first place can filled with $51$ characters, the second can be filled with only $50$ characters, ..., and finally the seventh place can be filled with the remaining $45$ characters.
So, the total number is
$$8\times 51\times 50\times \cdots \times 45\approx 4.7\times 10^{12}.$$

EDIT
There have been tow approaches to the same problem.
Problem
There are $n$ different symbols and $k$ slots to place them ($k<n$).
One of the symbols (the obligatory symbol) has a special role: it has to be part of all the arrangements. The other condition is that one symbol may appears only once.
Approach 1
One can place the obligatory symbol first. It may appear in $k$ different slots. As a result the the remaining $n-1$ symbols will have to be placed in the remaining $k-1$ slots. So the solution is
$$k\times(n-1)(n-2)\cdots (n-k+1).$$
If $n=52$ and $k=8$ then the result is
$$8\times 51\times 50\times 49 \times \cdots 45\approx 4.7\times 10^{12}$$
Approach 2
First we select $k-1$ symbols out of the $n-1$ ones and then multiply the resulting number with the possible orderings of the $k$ symbols (the obligatory one + the $k-1$ ones chosen.) The total is
$$\binom {n-1}{k-1}k!.$$
In our case the result is again $\binom{51}7\times 8!\approx 4.7\times 10^{12}$.
A: Yes you are correct.
Number of ways = $8 \times \frac{51!}{(51-7)!}$
