Exhaustive Guide to Factorials and Permutations

Properties of Factorials and Permutations - Second Secondary Grade

Introductory Information

• Subject: Algebra (فرع الجبر).
• Academic Level: Second Year of Secondary School (الصف الثاني الثانوي).
• Instructor: Osama Saad Allah (أسامة سعد الله).
• Website/Platform: OSAMASAADALLAIL.COM (referred to as SPICY).
• General Scope: Detailed study of factorials of positive integers and permutations.

Factorial of a Positive Integer ($n!$)

• Definition: The factorial of a positive integer $n$ is the product of all positive integers less than or equal to $n$.
• Mathematical Representation: It is written as $n!$ (or using the Arabic corner symbol).
• Formula:$n! = n imes (n-1) imes (n-2) imes \dots imes 3 imes 2 imes 1$
• Visualizing Factors:   - The number of factors in the expansion of $n!$ is exactly $n$ factors.   - The smallest factor is always $1$.   - The largest factor is always the number itself, $n$.

Core Properties of Factorials

• Base Case:   - $1! = 1$   - $0! = 1$ (By definition/convention).
• Equality with 1: If $n! = 1$, then the value of $n$ must be either $0$ or $1$.
• Recursive Expansion (The Extraction Rule):   - A factorial can be expressed in terms of a smaller factorial by "pulling out" the largest terms.   - $n! = n \times (n-1)!$   - $n! = n \times (n-1) \times (n-2)!$   - $n! = n \times (n-1) \times (n-2) \times (n-3)!$
• Divisibility: The factorial of any positive integer is divisible by the factorial of any positive integer smaller than it.   - $n!$ is divisible by $k!$ if $k \le n$.

Factorial Table of Common Values

• $1! = 1$
• $2! = 1 \times 2 = 2$
• $3! = 1 \times 2 \times 3 = 6$
• $4! = 1 \times 2 \times 3 \times 4 = 24$
• $5! = 1 \times 2 \times 3 \times 4 \times 5 = 120$
• $6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$

Permutations ($^n P_r$)

• Definition: Permutations represent the number of ways to arrange $r$ distinct items taken from a set of $n$ distinct items, where the order of arrangement is significant.
• Notation:   - English: $^n P_r$   - Arabic Context: $_n L_r$ (ن ل ر), where "ن" is the "Alam" (Total items) and "ر" is the "Dalil" (Items chosen).
• Constraints:   - $n$ must be a non-negative integer ($n \in \mathbb{Z}^+ \cup {0}$).   - $r$ must be a non-negative integer ($r \in \mathbb{Z}^+ \cup {0}$).   - $n \ge r$.
• The Sequential Multiplication Rule:   Permutations can be defined as the product of $r$ consecutive integers starting from $n$ and descending.   - $^n P_r = n \times (n-1) \times (n-2) \times \dots \times (n-r+1)$
• Permutations in Terms of Factorials:   - $^n P_r = \frac{n!}{(n-r)!}$

Special Cases of Permutations

• Ordering All Items: $^n P_n = n!$
• Ordering All But One: $^n P_{n-1} = n!$
• Zero Arrangements: $^n P_0 = 1$
• Ordering One Item: $^n P_1 = n$

Solved Examples: Factorials

Example 1: Basic Evaluation

• Question: Find $5!$ without a calculator.
• Solution: $5 \times 4 \times 3 \times 2 \times 1 = 120$.

Example 2: Fraction Simplification

• Question: Evaluate $\frac{9!}{8! + 7!}$.
• Solution:$\frac{9 \times 8 \times 7!}{(8 \times 7!) + (1 \times 7!)} = \frac{72 \times 7!}{(8+1) \times 7!} = \frac{72}{9} = 8$.

Example 3: Solving for Variable $n$

• Question: Find the value of $n$ if $\frac{(n+1)!}{(n-1)!} = 42$.
• Solution:   - Expand the numerator: $\frac{(n+1) \times n \times (n-1)!}{(n-1)!} = 42$   - Simplify: $(n+1) \times n = 42$   - Since $42 = 7 \times 6$, then $(n+1) = 7$ and $n = 6$.

Example 4: Finding $n$ from a Large Factorial

• Question: Find $n$ if $n! = 720$.
• Solution: Through successive division or knowledge of the factorial table:   - $720 / 6 = 120$   - $120 / 5 = 24$   - $24 / 4 = 6$   - $6 / 3 = 2$   - $2 / 2 = 1$   - Therefore, $n = 6$.

Example 5: Equation with Factorial Sums

• Question: Find $n$ if $\frac{(n+2)! + (n+1)!}{n!} = 56$.
• Solution:   - $\frac{(n+2)(n+1)n! + (n+1)n!}{n!} = 56$   - $(n+2)(n+1) + (n+1) = 56$   - Factor out $(n+1)$: $(n+1)[(n+2) + 1] = 56$   - $(n+1)(n+3) = 56$   - Finding two numbers with a difference of $2$ whose product is $56$: There are no such integers ($7 \times 8 = 56$ has a difference of $1$).   - Note: If the result were to be an integer (e.g., if the sum was $48$), $n=5$ would satisfy $(6)(8)=48$. Based on the slide handwritten note, assuming a typo and solving for the logic: $n=5$ results in $(6)(8) = 48$.

Solved Examples: Permutations

Example 1: Evaluation of Partial Arrangements

• Question: Evaluated $^5 P_2$.
• Solution: Start from $5$ and multiply two factors: $5 \times 4 = 20$.

Example 2: Permutation and Factorials

• Question: Evaluate $^9 P_3$.
• Solution: $9 \times 8 \times 7 = 504$.

Example 3: Finding Unknown in $^n P_r$

• Question: If $^n P_2 = 72$, find $n$.
• Solution:   - $n(n-1) = 72$   - Since $9 \times 8 = 72$, then $n = 9$.

Example 4: Solving for $r$

• Question: If $^6 P_r = 120$, find $r$.
• Solution:   - Start with $6$: $6 \times 5 = 30$ (two factors)   - $30 \times 4 = 120$ (three factors)   - Therefore, $r = 3$.

Example 5: Complex Algebra with Permutations

• Question: If $^n P_3 = 10 \times ^n P_2$, find $n$.
• Solution:   - Expand both sides: $n(n-1)(n-2) = 10 \times n(n-1)$   - Cancel $(n)(n-1)$ (where $n \neq 0, 1$): $n-2 = 10$   - $n = 12$.

Mathematical Challenges and Training

• Condition for Factorial Partition: $n! = k! \Rightarrow n = k$, or if $k! = 1$, then $n=0, 1$.
• Training Task 1: If $^n P_{n-2} = 60$, solve for $n$.
• Training Task 2: If $^n P_r = 504$, find $n$ and $r$.
• Training Task 3: If $^n P_r = 5040$, find $n$ and $r$.

Textbook References

• Page 45: Problems 1, 2, 3, 6, 8.
• Page 46: Problem 12.
• Page 48: Problems 35, 39, 48.