Application of FCP to Counting Problems Revision Notes for Grade 12 NSC Matric Mathematics

Application of FCP to Counting Problems

Counting problems are fundamental to probability theory and involve determining the number of ways to arrange or select objects under various conditions. These problems require systematic thinking and the application of core counting principles.

Understanding basic arrangements

When arranging objects in specific positions, we use the fundamental counting principle. This principle states that when making sequential choices, we multiply the number of available options at each step.

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Example: Race Positions

When arranging athletes in the first three places of a race with 8 competitors:

First place: 8 possible choices
Second place: 7 remaining choices
Third place: 6 remaining choices
Total arrangements: $8 \times 7 \times 6 = 336$ ways

This multiplication principle forms the foundation for solving more complex counting scenarios.

Arrangements with constraints

Many counting problems involve restrictions or conditions that certain objects must satisfy. These constraint problems require careful analysis of the given conditions.

Objects that must be together

When specific objects must remain adjacent or grouped together, we treat them as a single unit. This approach simplifies the problem considerably.

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Worked Example: Adjacent Seating

Seven boys of different ages must be seated on a bench, with the youngest and oldest boys sitting next to each other.

Solution approach:

Consider the youngest and oldest boys as one combined unit
We now have 6 objects to arrange: the combined unit plus 5 individual boys
The combined unit has $6!$ possible positions
Within the unit, the two boys can be arranged in $2!$ ways (youngest-oldest or oldest-youngest)
Total arrangements: $6! \times 2! = 720 \times 2 = 1440$ ways

Objects that must not be together

When certain objects cannot be adjacent, we use complementary counting: calculate all possible arrangements, then subtract the forbidden arrangements.

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Complementary Counting Strategy: Total valid arrangements = All possible arrangements - Forbidden arrangements

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Continuing the previous example: If the youngest and oldest boys must NOT sit next to each other:

Total unrestricted arrangements: $7! = 5040$
Arrangements where they sit together: $1440$ (calculated above)
Valid arrangements: $7! - 1440 = 3600$ ways

Arrangements with identical objects

When a set contains identical objects, the standard factorial formula overcounts arrangements. We must adjust by dividing by the factorial of each group of identical items.

Formula for repeated elements

For n objects where $n_1$ are identical, $n_2$ are identical, and so on:

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Formula for Identical Objects:

$\text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times ... \times n_k!}$

where $n$ is the total number of objects and $n_1, n_2, ..., n_k$ are the numbers of each type of identical object.

Letter arrangement examples

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Worked Example: Arranging "OMO"

The word contains 3 letters with two identical O's.

Treating all letters as distinct:

Total arrangements: $3! = 6$ possible arrangements

Treating identical letters as the same:

We divide by $2!$ to account for the repeated O's
Number of arrangements = $\frac{3!}{2!} = 3$
Valid arrangements: OMO, MOO, OOM

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Worked Example: Arranging "BASSOON"

This 7-letter word contains two S's and two O's.

All letters treated as different:

Total arrangements: $7! = 5040$

Repeated letters treated as identical:

Number of arrangements = $\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260$

Word must start with 'O':

Fix O in the first position
Arrange remaining 6 letters (including one O and two S's)
Number of arrangements = $\frac{6!}{2!} = 360$

Word must start and end with the same letter: Two possibilities exist:

Pattern S____S with middle letters B,A,S,O,O,N: $\frac{5!}{2!} = 60$ arrangements
Pattern O____O with middle letters B,A,S,S,O,N: $\frac{5!}{2!} = 60$ arrangements
Total: $60 + 60 = 120$ arrangements

Key problem-solving strategies

Understanding the right approach is crucial for solving counting problems efficiently. Here are the essential strategies:

The fundamental counting principle: Multiply the number of choices at each decision point when selections are independent and order matters.

Grouping strategy: When objects must stay together, treat the group as a single unit. Remember to account for internal arrangements within the group.

Complementary counting method: For "not together" or "exclusion" problems, calculate total arrangements minus unwanted arrangements.

Identical objects adjustment: When objects are indistinguishable, divide the total arrangements by the factorial of each group of identical items.

Worked example: Race arrangements

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Worked Example: Athletic Race Positions

Question: Eight athletes compete in a 400m race. In how many different ways can the first three places be arranged?

Solution: We need to fill three distinct positions from eight available athletes.

First place: 8 choices
Second place: 7 remaining choices
Third place: 6 remaining choices

Using the fundamental counting principle: Total arrangements = $8 \times 7 \times 6 = 336$ ways

Worked example: Seating with constraints

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Worked Example: Constrained Seating

Question: Seven boys must be seated on a bench. The youngest and oldest boys must sit next to each other. How many seating arrangements are possible?

Solution:

Step 1: Treat the youngest and oldest boys as a single unit

Step 2: We now have 6 objects to arrange (the pair plus 5 individuals)

Step 3: These 6 objects can be arranged in $6!$ ways

Step 4: Within their unit, the youngest and oldest can be arranged in $2!$ ways

Total arrangements = $6! \times 2! = 720 \times 2 = 1440$ ways

Worked example: Three-digit numbers

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Worked Example: Digit Arrangements

Question: How many three-digit numbers can be formed using digits 1 to 6 if repetition is not allowed?

Solution:

Hundreds place: 6 choices
Tens place: 5 remaining choices
Units place: 4 remaining choices

Total numbers = $6 \times 5 \times 4 = 120$

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Common Exam Mistakes to Avoid:

Forgetting internal arrangements: When grouping objects together, multiply by the factorial of objects within the group
Missing identical object adjustments: Always check for repeated elements before applying standard factorials
Misreading constraint requirements: Distinguish carefully between "must be together" and "must not be together" conditions
Complementary counting errors: Ensure both total and unwanted arrangements are calculated correctly
Order significance: Remember that arrangements consider position order, unlike combinations

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Key Points to Remember:

Fundamental counting principle: Multiply options at each step when choices are sequential and independent
Constraint handling: Group adjacent objects as units; use complementary counting for separation requirements
Identical objects formula: Always divide by factorials of repeated elements to avoid overcounting
Systematic approach: Break complex problems into manageable steps and check your logic
Practice recognition: Success comes from identifying which strategy applies to each problem type

Application of FCP to Counting Problems (Grade 12 NSC Matric Mathematics): Revision Notes