Possibilities for passwords with at least one lowercase and one uppercase letter

Question

Fred needs to choose a password for a certain website. Assume that he will choose an 8-character password, and that the legal characters are the lowercase letters a, b, c, . . . , z, the uppercase letters A, B, C, . . . , Z, and the numbers 0, 1, . . . , 9. (a) How many possibilities are there if he is required to have at least one lowercase letter in his password? (b) How many possibilities are there if he is required to have at least one lowercase letter and at least one uppercase letter in his password? (c) How many possibilities are there if he is required to have at least one lowercase letter, at least one uppercase letter, and at least one number in his password? (Introduction to Probability, Blitzstein and Nwang, p.38)

Part a) is no problem, but I don't know how to define the events for b) and c). Any hint for defining the events to be able to use inclusion-exclusion?

EDIT1

Here is how I did part a).

The question is how many ways are there to choose a password having at least one lowercase letter.

• Define $A_i$ as the set of all passwords with lowercase letter at the $i^{th}$ position.
• $|A_i| = 26^1 \; 62^{8-1}$
• $|A_i \cap A_j| = 26^2 \; 62^{8-2}$
• $|\cap_{i=1}^8 A_i| = 26^i \; 62^{8-i}$

\begin{equation} |\cup_{i=1}^8 A_i| = \sum_{i=1}^8 \binom{8}{i} 26^i \; 62^{8-i} (-1)^{i+1} \end{equation}

I would like to solve the other parts in exactly the same manner, however, I don't know how to define my sets $A_i$ etc. in this case.

EDIT2

I try part c) with @AlexR's approach:

$|C| = |\Omega| - (|\neg U| + |\neg L| + |\neg N|) + (|\neg U \cap \neg L| + |\neg U \cap \neg N| + |\neg L \cap \neg N|) - |\neg U \cap \neg L \cap \neg N| = 62^8 - (36^8 + 36^8 + 52^8) + (10^8 + 26^8 + 26^8) - 0^8$

The probability of this to happen would be $0.7312$.

Answer 1

(b): How many passwords are there minus how many passwords are there with no uppercase or no lowercase letter?
(c): How many passwords are there minus how many passwords are there with no uppercase or no lowercase letter or no number?

Notation: $$\Omega := \{ p\ |\ p\text{ is an } 8 \text{-digit password from the alphabet } [A-Z,a-z, 0-9]\}\\ U := \{p \in\Omega\ |\ p\text{ contains an uppercase letter}\} \\ L := \{p \in\Omega\ |\ p\text{ contains a lowercase letter}\} \\ \neg U := \Omega \setminus U = \{p\ |\ p\text{ is an } 8 \text{-digit password from the alphabet } [a-z,0-9]\} \\ \neg L := \Omega \setminus L = \{p\ |\ p\text{ is an } 8 \text{-digit password from the alphabet } [A-Z,0-9]\}$$ We trivially have $$|\Omega| = 62^8, |\neg U| = |\neg L| = 36^8$$

Example for (a):
$$|A| = |L| = |\Omega| - |\neg L| = 62^8 - 36^8$$

Example for (b):
Now there are $36$ non-uppercase and $36$ non-lowercase characters as well as $10$ non-letters so $$|B| = |U\cap L| = |\Omega| - |\neg U| - |\neg L| + |\neg U \cap \neg L| = 62^8 - 36^8 - 36^8 + 10^8$$ by inclusion-exclusion.

Can you do (c) now?

Answer 2

Suggestion: consider the multinomial coefficients in multi-index notation. I hope what I am doing below is correct!

a) $$\sum_{j\geq 1} {8\choose j}26^j 36^{8-j}=\underset{\alpha_1\geq 1\\|\alpha|=8}{\sum} {8\choose \alpha}26^{\alpha_1} 26^{\alpha_2}10^{\alpha_3},$$ where $\alpha\in\mathbb N_0^3$ ranges over multiindices $\alpha=(\alpha_1,\alpha_2,\alpha_3)$ with $\alpha_1\geq 1$ and $|\alpha|=\alpha_1+\alpha_2+\alpha_3=8$ by definition. Here, the counting goes as follows: there are ${8\choose \alpha}$ ways to pick $\alpha_1$ slots in which to put lowercase letters, $\alpha_2$ slots in which to put upper case letters and $\alpha_3$ slots in which to put numbers. For any single such choice, there is $26^{\alpha_1} 26^{\alpha_2}10^{\alpha_3}$ ways to put $j$ lowercase letters in the $\alpha_1$ chosen lower case slots, $\alpha_2$ upper case case letters in the chosen upper case slots and $\alpha_3$ numbers in the chosen number slots.

The sum can be evaluated by recalling that $$\underset{|\alpha|=8\\\alpha\in \mathbb N_0^d}{\sum}{8\choose \alpha}x^{\alpha}=(x_1+\ldots+x_d)^8.$$ In particular $$ \underset{\alpha_1\geq 1\\|\alpha|=8}{\sum} {8\choose \alpha}26^{\alpha_1} 26^{\alpha_2}10^{\alpha_3}=\underset{|\alpha|=8}{\sum} {8\choose \alpha}26^{\alpha_1} 26^{\alpha_2}10^{\alpha_3}-\underset{\alpha_1=0\\|\alpha|=8}{\sum} {8\choose \alpha} 26^{\alpha_2}10^{\alpha_3}=62^8-36^8 $$

b) \begin{align*}\underset{\alpha_1\geq 1\\\alpha_2\geq 1\\|\alpha|=8}{\sum} {8\choose \alpha}26^{\alpha_1} 26^{\alpha_2} 10^{\alpha_3}&=-\underset{|\alpha|=8}{\sum} {8\choose \alpha}26^{\alpha_1} 26^{\alpha_2} 10^{\alpha_3}+2\underset{\alpha_1\geq 1\\|\alpha|=8}{\sum} {8\choose \alpha}26^{\alpha_1} 26^{\alpha_2} 10^{\alpha_3}\\ &\hspace{2cm}+\underset{\alpha_1=0\\\alpha_2=0\\|\alpha|=8}{\sum} {8\choose \alpha}26^{\alpha_1} 26^{\alpha_2} 10^{\alpha_3}=-62^8+2(62^8-36^8)+10^8. \end{align*}

c) $$\underset{\alpha_1\geq 1\\\alpha_2\geq 1\\\alpha_3\geq 1\\|\alpha|=8}{\sum} {8\choose \alpha}26^{\alpha_1} 26^{\alpha_2} 10^{\alpha_3}.$$ In order to evaluate the sum, we can evoke the inclusion-exclusion principle for sums. Let $\Omega$ denote those multiindeces in $\mathbb N_0^3$ for which $|\alpha|=8$. Then we have $$\Omega=\{\alpha_1=0\}\cup \{\alpha_2=0\}\cup \{\alpha_3=0\}\cup \{\alpha_1,\alpha_2,\alpha_3\geq 1\}.$$ We have to perform the sum over elements from $\{\alpha_1,\alpha_2,\alpha_3\geq 1\}$. Accordingly, we use the analog of the inclusion-exclusion principle for sums to arrive at $$ \sum_{|\alpha|=8}=\sum_{\alpha_1=0}+\sum_{\alpha_2=0} + \sum_{\alpha_3=0}+\sum_{\alpha_1,\alpha_2,\alpha_3\geq 1}-\sum_{\alpha_1=\alpha_2=0}-\sum_{\alpha_2=\alpha_3=0}-\sum_{\alpha_3=\alpha_1=0}+\sum_{\alpha_1=\alpha_2=\alpha_3=0}. $$ A different approach, more akin to what we did in b), would be to use $$ \Omega=\{\alpha_1\geq 1\}\cup \{\alpha_2\geq 1\}\cup \{\alpha_3\geq 1\}. $$ The result is the same, but the calculations are more daunting.