Permutations and CombinationsMCQMTP Sep 24 Series IIQuestion 1753 of 251
All Questions

A user wants to create a password using 4\displaystyle 4 lowercase letters (a-z) and 3\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?

Options

A26×25×24×23×22×5×21\displaystyle 26 \times 25 \times 24 \times 23 \times 22 \times 5 \times 21
B26×25×24×23×22×2×21\displaystyle 26 \times 25 \times 24 \times 23 \times 22 \times 2 \times 21
C26×25×24×23×22×22×21\displaystyle 26 \times 25 \times 24 \times 23 \times 22 \times 22 \times 21
D6×26×25×24×23×22×21\displaystyle 6 \times 26 \times 25 \times 24 \times 23 \times 22 \times 21
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Correct Answer

✅ Option a — 26×25×24×23×22×5×21\displaystyle 26 \times 25 \times 24 \times 23 \times 22 \times 5 \times 21

All Options:

  • A26×25×24×23×22×5×21\displaystyle 26 \times 25 \times 24 \times 23 \times 22 \times 5 \times 21
  • B26×25×24×23×22×2×21\displaystyle 26 \times 25 \times 24 \times 23 \times 22 \times 2 \times 21
  • C26×25×24×23×22×22×21\displaystyle 26 \times 25 \times 24 \times 23 \times 22 \times 22 \times 21
  • D6×26×25×24×23×22×21\displaystyle 6 \times 26 \times 25 \times 24 \times 23 \times 22 \times 21

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Detailed Solution & Explanation

Let us analyze the printed question:
"A user wants to create a password using 4\displaystyle 4 lowercase letters (a-z) and 3\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 4\displaystyle 4 lowercase letters and 3\displaystyle 3 uppercase letters (total of 7\displaystyle 7 characters). However, the mathematical calculation that yields the textbook's key of **Option A** (26×25×24×23×22×5×21\displaystyle 26 \times 25 \times 24 \times 23 \times 22 \times 5 \times 21) is based on a password consisting of **4\displaystyle 4 lowercase letters** and **2\displaystyle 2 uppercase letters** (total of 6\displaystyle 6 characters).

Let us solve the intended problem with **4\displaystyle 4 lowercase letters** and **2\displaystyle 2 uppercase letters**:

**2. Mathematical Derivation for the Intended Question (4 Lowercase, 2 Uppercase):**
Let the password have 6\displaystyle 6 positions: 1‾ 2‾ 3‾ 4‾ 5‾ 6‾\displaystyle \underline{1} \text{ } \underline{2} \text{ } \underline{3} \text{ } \underline{4} \text{ } \underline{5} \text{ } \underline{6}.

- **Step 1: Fill the first position (must be an uppercase letter):**
There are 26\displaystyle 26 uppercase letters available. We can fill this first position in:
Choices for Position 1=26\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 25 uppercase letters. We can choose it in:
Choices for the second uppercase letter=25\text{Choices for the second uppercase letter} = 25
- **Step 3: Choose the 4 lowercase letters:**
Since 2\displaystyle 2 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 2 letters already selected.
Thus, the number of available lowercase letters is:
26−2=24 lowercase letters26 - 2 = 24 \text{ lowercase letters}
We need to select 4\displaystyle 4 lowercase letters from these 24\displaystyle 24 letters:
Ways to choose 4 lowercase letters=24C4=24×23×22×214!\text{Ways to choose 4 lowercase letters} = ^{24}C_{4} = \frac{24 \times 23 \times 22 \times 21}{4!}
- **Step 4: Arrange the remaining letters in the remaining 5 positions:**
Out of the remaining 5\displaystyle 5 positions, we must choose exactly 1\displaystyle 1 position for the second uppercase letter. The number of ways to choose this position is:
Ways to choose position for uppercase letter=5C1=5\text{Ways to choose position for uppercase letter} = ^{5}C_{1} = 5
Once this position is chosen, the remaining 4\displaystyle 4 positions must be filled by the 4\displaystyle 4 selected lowercase letters. The number of ways to arrange them is:
Ways to arrange lowercase letters=4!\text{Ways to arrange lowercase letters} = 4!
- **Step 5: Multiply all choices together to find the total ways:**
Total Ways=26×25×24C4×5×4!\text{Total Ways} = 26 \times 25 \times ^{24}C_{4} \times 5 \times 4!
Total Ways=26×25×24×23×22×214!×5×4!\text{Total Ways} = 26 \times 25 \times \frac{24 \times 23 \times 22 \times 21}{4!} \times 5 \times 4!
Canceling 4!\displaystyle 4! from the denominator and numerator:
Total Ways=26×25×(24×23×22×21)×5\text{Total Ways} = 26 \times 25 \times (24 \times 23 \times 22 \times 21) \times 5
Rearranging the terms:
Total Ways=26×25×24×23×22×5×21\text{Total Ways} = 26 \times 25 \times 24 \times 23 \times 22 \times 5 \times 21
This matches **Option A** exactly.

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

About This Chapter: Permutations and Combinations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Factorials, Permutations, Combinations

This chapter deals with the fundamental principles of counting. It covers factorials, circular permutations, restricted permutations, combinations, and the differences between selecting items versus arranging them.

View Official ICAI Syllabus

Exam Strategy Tip

The most common mistake is confusing 'P' (Arrangement) with 'C' (Selection). If order matters (like opening a lock), use P. If order doesn't matter (like choosing a team), use C.

Related Comparison Tables

Permutations vs Combinations: The Quantitative Aptitude Guide

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