A user wants to create a password using $\displaystyle 4$ lowercase letters (a-z) and $\displaystyle 3$ uppercase letters (A-Z). No letter can be repeated in any form. In how many ways can the password be created if the password must start with an uppercase letter?

Let us analyze the printed question:
"A user wants to create a password using $\displaystyle 4$ lowercase letters (a-z) and $\displaystyle 3$ uppercase letters (A-Z). No letter can be repeated in any form. In how many ways can the password be created if the password must start with an uppercase letter?"

**1. Identifying the Typographical Error in the Question:**
The question text states that the password uses $\displaystyle 4$ lowercase letters and $\displaystyle 3$ uppercase letters (total of $\displaystyle 7$ characters). However, the mathematical calculation that yields the textbook's key of **Option A** ($26 $\displaystyle \times$ 25 $\displaystyle \times$ 24 $\displaystyle \times$ 23 $\displaystyle \times$ 22 $\displaystyle \times$ 5 $\displaystyle \times$ 21$\displaystyle ) is based on a password consisting of **$4$\displaystyle lowercase letters** and **$2$\displaystyle uppercase letters** (total of$6$\displaystyle characters).<br><br>Let us solve the intended problem with **$4$\displaystyle lowercase letters** and **$2$\displaystyle uppercase letters**:<br><br>**2. Mathematical Derivation for the Intended Question (4 Lowercase, 2 Uppercase):**<br>Let the password have$6$\displaystyle positions:$$\displaystyle \underline{1}$ $\displaystyle \text{ }$ $\displaystyle \underline{2}$ $\displaystyle \text{ }$ $\displaystyle \underline{3}$ $\displaystyle \text{ }$ $\displaystyle \underline{4}$ $\displaystyle \text{ }$ $\displaystyle \underline{5}$ $\displaystyle \text{ }$ $\displaystyle \underline{6}<span$ class="katex-error" title="ParseError: KaTeX parse error: Expected 'EOF', got '&' at position 106: …*<br>There are &̲#36;26" style="color:#cc0000">\displaystyle .<br><br>- **Step 1: Fill the first position (must be an uppercase letter):**<br>There are &#36;26 uppercase letters available. We can fill this first position in:
$$\text{Choices for Position 1}=26$$$\displaystyle \text{Choices for Position 1} = 26$
- **Step 2: Choose the second uppercase letter:**
Since no letter can be repeated in any form, the second uppercase letter must be chosen from the remaining $25$\displaystyle uppercase letters. We can choose it in:<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mtext>Choices for the second uppercase letter</mtext><mo>=</mo><mn>25</mn></mrow><annotation encoding="application/x-tex">$$\displaystyle \text{Choices for the second uppercase letter} = 25$</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8889em;vertical-align:-0.1944em;"></span><span class="mord text"><span class="mord">Choices for the second uppercase letter</span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">25</span></span></span></span></span><br>- **Step 3: Choose the 4 lowercase letters:**<br>Since2$ uppercase letters have already been chosen, and no letter can be repeated in any form (meaning the lowercase and uppercase versions of the same letter are excluded), we cannot use the lowercase versions of the $2$\displaystyle letters already selected.<br>Thus, the number of available lowercase letters is:<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mn>26</mn><mo>−</mo><mn>2</mn><mo>=</mo><mn>24</mn><mtext> lowercase letters</mtext></mrow><annotation encoding="application/x-tex">26 - 2 = 24$$\displaystyle \text{ lowercase letters}$</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7278em;vertical-align:-0.0833em;"></span><span class="mord">26</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">2</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord">24</span><span class="mord text"><span class="mord"> lowercase letters</span></span></span></span></span></span><br>We need to select4$ lowercase letters from these $24$ letters:
$$\text{Ways to choose 4 lowercase letters}{=}^{24}{C}_4=\frac{24\times 23\times 22\times 21}{4!}$$\displaystyle \text{Ways to choose 4 lowercase letters} = ^{24}C_{4} = \frac{24 $\displaystyle \times$ 23 $\displaystyle \times$ 22 $\displaystyle \times$ 21}{4!}
- **Step 4: Arrange the remaining letters in the remaining 5 positions:**
Out of the remaining $\displaystyle 5$ positions, we must choose exactly $\displaystyle 1$ position for the second uppercase letter. The number of ways to choose this position is:
$$\text{Ways to choose position for uppercase letter}{=}^5{C}_1=5$$$\displaystyle \text{Ways to choose position for uppercase letter} = ^{5}C_{1} = 5$
Once this position is chosen, the remaining $\displaystyle 4$ positions must be filled by the $\displaystyle 4$ selected lowercase letters. The number of ways to arrange them is:
$$\text{Ways to arrange lowercase letters}=4!$$$\displaystyle \text{Ways to arrange lowercase letters} = 4$!
- **Step 5: Multiply all choices together to find the total ways:**
$$\text{Total Ways}=26\times 25{\times }^{24}{C}_4\times 5\times 4!$$$\displaystyle \text{Total Ways} = 26$ $\displaystyle \times$ 25 $\displaystyle \times$ ^{24}C_{4} $\displaystyle \times$ 5 $\displaystyle \times$ 4!
$$\text{Total Ways}=26\times 25\times \frac{24\times 23\times 22\times 21}{4!}\times 5\times 4!$$$\displaystyle \text{Total Ways} = 26$ $\displaystyle \times$ 25 $\displaystyle \times$ $\displaystyle \frac{24 \times 23 \times 22 \times 21}{4!}$ $\displaystyle \times$ 5 $\displaystyle \times$ 4!
Canceling $4!$ from the denominator and numerator:
$$\text{Total Ways}=26\times 25\times (24\times 23\times 22\times 21)\times 5$$$\displaystyle \text{Total Ways} = 26$ $\displaystyle \times$ 25 $\displaystyle \times$ (24 $\displaystyle \times$ 23 $\displaystyle \times$ 22 $\displaystyle \times$ 21) $\displaystyle \times$ 5
Rearranging the terms:
$$\text{Total Ways}=26\times 25\times 24\times 23\times 22\times 5\times 21$$$\displaystyle \text{Total Ways} = 26$ $\displaystyle \times$ 25 $\displaystyle \times$ 24 $\displaystyle \times$ 23 $\displaystyle \times$ 22 $\displaystyle \times$ 5 $\displaystyle \times$ 21
This matches **Option A** exactly.

Hence, **Option A** is the correct answer.