How many passwords with 8 characters have :

Question

How many passwords with $8$ characters ( $26$ lowercase $26$ uppercase and $10$ digits ) have :

a) exactly $3$ lowercase characters $3$ uppercase characters and $2$ digits

b) all characters are different

for $b)$ the answer should simply be $\binom{62}{8}$ because we need different characters.

for $a$) there should be some multinomial distribution right there. We can have $\frac{8!}{3!3!2!}$ ways to get the $ 3-3-2 $ distribution but we need to choose from the total number of lowercase/uppercase characters and digits. Do we simply multiply by the binomial coefficient? How to approach this?

Answer 1

a)

For an ordered sequence of $3$ lowercase letters there are $26^3$ possibilities.

For an ordered sequence of $3$ uppercase letters there are $26^3$ possibilities.

For an ordered sequence of $2$ digits there are $10^2$ possibilities.

Maintaining the order of these sequences we must select $3$ of $8$ spots for the lowercase letters, $3$ of $8$ spots for the uppercase letters and $2$ of $8$ spots for the digits, and this gives $\frac{8!}{3!3!2!}$ possibilities.

We conclude that there are: $$\frac{8!}{3!3!2!}\times 26^3\times26^3\times10^2=\frac{8!}{3!3!2!}\times 26^6\times10^2$$such passwords.

b)

There are $26+26+10=62$ possibilities for the first character, $61$ for the second character, et cetera.

So there are $$62\times61\times\cdots\times55=\frac{62!}{54!}$$such passwords.

Answer 2

a) First choose the positions of the digits, then assign the digits, then choose the positions of the uppercase letters, and assign the letters. The total number of combinations thus equals:

$${8 \choose 2} 10^2 {6 \choose 3} 26^6$$

b) First choose the characters, then assign them to their position in the sequence:

$${62 \choose 8} 8!$$