How many ways of creating a password of length 7 How many ways of creating a password of length $7$ given constraint:
$5$ upper case, $1$ lower case, and $1$ digit.
My answer is there are $^7C_{5}$ way of choosing position for uppercase letter, and for each one of the $^7C_{5}$ way we have $26^5$ ways of choosing a password as we have $26$ alphabets.
For each of $^7C_{5}$ x $26^5$ ways of choosing above, we can pick $^2C_{1}$ position in the password string to place a lowercase, and for each one of the $^2C_{1}$ ways we have $26$ way of choosing lowercase.
For each of $^7C_5$ x $26^5$ x $^2C_1$ x $26$ ways of choosing upper case and lower cases, we have $10$ options to pick digit for the very last position left over. 
Then my answer is : $^7C_5$ x $26^5$ x $^2C_1$ x $26$ x $10$ passwords can be created.
Please tell me how I got it wrong?
 A: Your answer is correct.
There are $7$ ways to choose the position of the lower case letter and $26$ ways to fill that position with a lower case letter.  There are $6$ positions remaining.  Choose one of them to fill with a digit.  That digit may be chosen in $10$ ways.  The remaining five positions must be filled with upper case letters.  Each of those positions may be filled in $26$ ways.  Hence, there are 
$$7 \cdot 26 \cdot 6 \cdot 10 \cdot 26^5 = 7 \cdot 6 \cdot 10 \cdot 26^6$$
admissible passwords, as you found.
A: We should choose the element which wee include in the password
Eg:a letter from the alphabet or  a digit form 0-9;
Then we consider the permutations including those elements.
Problem:
Choosing a lowercase letter = $26\choose 1$ = $26$
Choosing a digit = $10\choose1$ = $10$
Problem occurs when choosing the uppercase letters.
Consider the number of total combinations in a case-wise manner
Case 1: All distinct letters : ${26 \choose 5} * 26 * 10$
Case 2: $2$ identical letters and $3$ distinct letters= $ {26 \choose 1} * {25 \choose 3} * 26 * 10$ 
Case 3: $3$ identical letters and $2$ distinct letters= ${26 \choose 1} * {25\choose 2} * 26 * 10$
Case 4: $4$ identical letters and $1$ distinct letters= ${26 \choose 1} * {25 \choose 1} * 26 * 10$
Case 5: All identical letters= ${26 \choose 1} *26 * 10$ 
Let us denote $P_i$ to the number of permutations for the $i^{th}$ case.
Thereby;
$$P_1 = {26 \choose 5} * 7!* 26 * 10$$
$$P_2 = {26 \choose 1} * {25 \choose 3} * 26 * 10* \frac{7!}{2!}$$
$$P_3 = {26 \choose 1} * {25\choose 2} * 26 * 10* \frac{7!}{3!} $$
$$P_4 = {26 \choose 1} * {25 \choose 1} * 26 * 10* \frac{7!}{4!} $$
$$P_5 = {26 \choose 1} * 26 * 10* \frac{7!}{5!}$$
Therefore $$\sum_{i=1}^5 P_i $$ would give the total number of passwords possible.
