Calculate the possible combinations for an eight-character password Can any when one help me in this question: 
Calculate the possible combinations for;


*

*An eight-character password consisting of upper- and lowercase letters and at least one numeric digit $(0–9)$?

*A ten-character password consisting of upper- and lowercase letters and at least one numeric digit $(0–9)$?
Thanks 
 A: Hint:
(v): One character can be one of 62 (= 26[A-Z]+26[a-z]+10[0-9]) letters, and choosing one character for password is independent of choosing other characters for password.
(vi): #(An alphanumeric password containing at least one numeric digit) = #(An alphanumeric password) - #(An alphanumeric password which not contains any digit) 
A: For both cases, you need to find the total number of combinations and then subtract the number of ones that don't match.
For reference, there are 26 capital letters, 26 lowercase letters, and 10 digits.
For the first case, let $N_8$ be the total number of 8 character passwords and $W_8$ be those without digits.
$$
N_8 = \binom{26+26+10}{1}^8 = \binom{62}{1}^8 = 62^8 \\
X_8 = \binom{26+26}{1}^8 =\binom{52}{1}^8 = 52^8
$$
Now let $D_8$ be the number of 8 character passwords with at least one digit.
$$
D_8 = N_8 - X_8 = 62^8 - 52^8 \approx 1.64 * 10^{14}
$$
Do the same for the second case:
$$
N_{10} = \binom{62}{1}^{10} = 62^{10} \\
X_{10} = \binom{52}{1}^{10} = 52^{10} \\
D_{10} = N_{10} - X_{10} = 62^{10} - 52^{10} \approx 6.95 *10^{17}
$$
A: assuming repetition allowed for an eight digit password 
you need one lower case which is $^{26}C_1$ and you need one upper case which is $^{26}C_1$ and you need one digit which is $^{10}C_1$ 
so we got $3$ we need another $5$ letters or numbers and repetition allowed and we have $62$ choices for each of the $5$ remaining characters 
so $\space ^{26}C_1 \cdot \ ^{26}C_1 \cdot \ ^{10}C_1\cdot \ (^{62}C_1)^5$
so for a $10$ letter password same thing first $3$ steps which is you need one lower case which is $^{26}C_1$ and you need one upper case which is $^{26}C_1$ and you need one digit which is $^{10}C_1 $
so we got $3$ we need another $7$ letters or numbers and repetition allowed and we have $62$ choices for each of the $7$ remaining characters 
so $ \ ^{26}C_1 \cdot \ ^{26}C_1 \cdot \ ^{10}C_1 \cdot \ (^{62}C_1)^7$
A: If you are using upper case thats $26$ the add $26$ for lower case and thats $52$. Now there are also $10$ numbers, thats $62$. Typically if we are talking passwords there are only $8$ special characters, so there are $70$ possibilities for the $8$ difference digits. $70 \times 8 = 560$.
Now abcdefgh, no there are $8$ characters. to figure out how many different possibilities for combinations there are thats $8!= 40,320$.
Here is where I am a little lost, if I wanted to include all $70$ options for all $8$ digits is it $560 + 40,320$? or $560 \times 40,320$?
