Chapter 6 — PERMUTATIONS AND COMBINATIONS

Theory

This problem is based on the Fundamental Principle of Counting (FPC): If a task can be performed in multiple independent steps, then total ways = product of number of ways of each step.

A 3-digit number consists of three positions: Hundreds, Tens, Units. Each position is filled according to given constraints.

Solution Roadmap

Step 1: Identify number of positions (3-digit ⇒ 3 places)
Step 2: Check restriction (repetition allowed or not)
Step 3: Assign choices for each position
Step 4: Multiply using FPC

Solution

Given digits: \(1,2,3,4,5\)

Case (i): Repetition Allowed

Each position (H, T, U) can be filled in 5 ways.

\[ \begin{aligned} \text{Total numbers} &= 5 \times 5 \times 5 \ &= 125 \end{aligned} \]

Case (ii): Repetition Not Allowed

First position: 5 choices
Second position: 4 remaining choices
Third position: 3 remaining choices

\[ \begin{aligned} \text{Total numbers} &= 5 \times 4 \times 3 \ &= 60 \end{aligned} \]

Final Answer

(i) \(125\)
(ii) \(60\)

Exam Significance

This is a foundation problem for permutations and combinations. It directly builds intuition for:

• Digit formation problems (very common in boards)
• JEE pattern questions involving restrictions
• Understanding when order matters vs selection

In competitive exams, this concept is frequently extended to: numbers with conditions, divisibility, repetition constraints.

Theory

A number is even if its unit digit is divisible by 2. Therefore, the unit place must contain one of the even digits.

This problem again uses the Fundamental Principle of Counting (FPC), but with a restriction on the last digit.

Solution Roadmap

Step 1: Identify positions → Hundreds, Tens, Units
Step 2: Apply condition → Unit must be even
Step 3: Count choices for each position
Step 4: Multiply using FPC

Solution

Given digits: \(1,2,3,4,5,6\)

Step 1: Unit place (even condition)
Even digits = \(2,4,6\) ⇒ 3 choices

Step 2: Hundreds place
Any digit can be used ⇒ 6 choices

Step 3: Tens place
Repetition allowed ⇒ 6 choices

\[ \begin{aligned} \text{Total numbers} &= 6 \times 6 \times 3 \ &= 108 \end{aligned} \]

Final Answer

\(108\)

Exam Significance

This is a classic constraint-based counting problem. It strengthens your ability to:

• Handle digit restrictions (even, odd, divisible numbers)
• Apply condition on a specific position (units place control)
• Use FPC under constraints

In Boards, similar direct questions are frequently asked.
In JEE/competitive exams, this idea is extended to: divisibility, sum constraints, and digit repetition patterns.

Theory

When order matters and repetition is not allowed, the situation corresponds to Permutations.

The number of ways to arrange \(r\) objects out of \(n\) distinct objects is:

\[ ^nP_r = \frac{n!}{(n-r)!} \]

Here, we are forming 4-letter codes from 10 distinct letters without repetition, so this is a direct application of permutation.

Solution Roadmap

Step 1: Identify total letters → 10
Step 2: Number of positions → 4
Step 3: No repetition ⇒ choices decrease each step
Step 4: Apply FPC or permutation formula

Solution

Total letters available = 10

Since repetition is not allowed:

First position: 10 choices
Second position: 9 choices
Third position: 8 choices
Fourth position: 7 choices

\[ \begin{aligned} \text{Total codes} &= 10 \times 9 \times 8 \times 7 \ &= 5040 \end{aligned} \]

\[ \text{Also, } ^{10}P_4 = \frac{10!}{6!} = 5040 \]

Final Answer

\(5040\)

Exam Significance

This is a direct permutation model problem. It builds the bridge between: Fundamental Counting → Permutation Formula.

• Very common in CBSE board exams (direct or slight variation)
• Core pattern for JEE questions involving arrangements
• Used in coding, password, and arrangement-based problems

In competitive exams, this concept extends to: restricted permutations, circular arrangements, and position constraints.

Theory

This problem combines: fixed positions + permutations without repetition.

When some positions are fixed, we only count arrangements of the remaining positions. Since repetition is not allowed, choices reduce at each step.

This can also be viewed as: selecting and arranging remaining digits ⇒ permutation of remaining digits.

Solution Roadmap

Step 1: Fix given digits (67)
Step 2: Identify remaining positions (3 places)
Step 3: Count remaining available digits
Step 4: Apply decreasing choices (no repetition)

Solution

Total digits available = \(0,1,2,3,4,5,6,7,8,9\) ⇒ 10 digits

First two digits are fixed as \(6\) and \(7\) ⇒ 1 way each

Remaining digits after using 6 and 7: \(0,1,2,3,4,5,8,9\) ⇒ 8 digits

Now fill remaining 3 positions:

Third position: 8 choices
Fourth position: 7 choices
Fifth position: 6 choices

\[ \begin{aligned} \text{Total numbers} &= 1 \times 1 \times 8 \times 7 \times 6 \ &= 336 \end{aligned} \]

\[ \text{Also, } ^8P_3 = 8 \times 7 \times 6 = 336 \]

Final Answer

\(336\)

Exam Significance

This is a high-value pattern problem frequently seen in exams. It builds mastery in:

• Handling fixed digit constraints
• Combining restriction + permutation
• Reducing problem size before applying formulas

In Boards, this appears in direct form.
In JEE/competitive exams, it evolves into: partial restrictions, position locking, and conditional permutations.

Theory

Each coin toss has 2 possible outcomes: Head (H) or Tail (T).

Since each toss is independent, the total number of outcomes is found using the Fundamental Principle of Counting (FPC).

This also forms the basis of sample space in probability, where all possible outcomes are counted.

Solution Roadmap

Step 1: Identify number of trials → 3 tosses
Step 2: Outcomes per trial → 2 (H or T)
Step 3: Multiply using FPC
Step 4: (Optional) Visualize sample space

Solution

Each toss has 2 possible outcomes.

For 3 tosses:

\[ \begin{aligned} \text{Total outcomes} &= 2 \times 2 \times 2 \ &= 2^3 \ &= 8 \end{aligned} \]

The sample space is: \[ \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \]

Final Answer

\(8\)

Exam Significance

This is a bridge problem between counting and probability.

• Forms the base of sample space questions in probability
• Frequently asked in CBSE exams (direct or with variation)
• Essential for JEE probability problems

In advanced problems, this concept extends to: binomial distribution, probability of events, and conditional probability.

Theory

When order matters, the problem is treated as a permutation.

Here, "one below the other" means: changing the order changes the signal.

Example: (Red above Blue) ≠ (Blue above Red)

Hence, we are arranging 2 flags out of 5 ⇒ permutation:

\[ ^5P_2 = \frac{5!}{3!} \]

Solution Roadmap

Step 1: Identify positions → Top and Bottom
Step 2: Check if order matters → Yes
Step 3: Apply decreasing choices (no repetition)
Step 4: Multiply using FPC or use permutation formula

Solution

Total flags = 5 (all distinct)

Top position: 5 choices
Bottom position: 4 remaining choices

\[ \begin{aligned} \text{Total signals} &= 5 \times 4 \ &= 20 \end{aligned} \]

\[ \text{Also, } ^5P_2 = 5 \times 4 = 20 \]

Final Answer

\(20\)

Exam Significance

This is a core concept question that tests whether you can identify order vs selection.

• Very common confusion: permutation vs combination
• Directly asked in CBSE boards
• Highly important for JEE (basic screening concept)

Key takeaway: If arrangement changes meaning → use permutation
If only selection matters → use combination