Factorial

FACTORIAL

Chapter 7 of the sources is dedicated to the concept of Factorials, which serves as a vital mathematical tool for the principles of counting and probability discussed in the surrounding chapters.

Key Concepts and Formulas

💡

The Factorial Formula ( $n!$ )

The product of the first $n$ positive integers is called $n$ factorial.

$n! = n \times (n - 1) \times (n - 2) \times \dots \times 1$

💡

The Zero Convention

By mathematical convention, $0! = 1$ .

💡

The Recursive Property

For any integer $i \le n$ , you can “stop” the factorial expansion at any point by adding a factorial sign to the remaining product.

Formula: $n! = n \times (n - 1) \times \dots \times (n - i + 1) \times (n - i)!$ .
Example: $5! = 5 \times 4!$ or $5! = 5 \times 4 \times 3!$ .

Examples for Understanding

1. Applied Counting

In an 8-athlete race with no ties, the total possible ways they can finish is the product of available spots:

$8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \mathbf{8!}$

2. Simplifying Fractions

To simplify $\frac{6!}{3!}$ , you expand the larger factorial until it matches the smaller one, allowing them to cancel out:

$\frac{6 \times 5 \times 4 \times 3!}{3!} = 6 \times 5 \times 4 = \mathbf{120}$

3. Expressing Products as Factorials

To express $25 \times 24 \times 23$ as a factorial, you complete the sequence down to 1 and then divide by the part you added:

$\frac{25 \times 24 \times 23 \times (22 \times 21 \times \dots \times 1)}{22 \times 21 \times \dots \times 1} = \mathbf{\frac{25!}{22!}}$

Practice Questions and Solutions

Solving for $n$

If $\frac{(n-1)!}{(n-3)!} = 6$ , find the value of $n$ .

View Detailed Solution ▼

Expand $(n-1)!$ until it reaches $(n-3)!$ : $\frac{(n-1)(n-2)(n-3)!}{(n-3)!} = 6 \implies (n-1)(n-2) = 6$ $n^2 - 3n + 2 = 6 \implies n^2 - 3n - 4 = 0$ $(n-4)(n+1) = 0$ Since $n$ must be a positive integer, $n = 4$ .

Expression Conversion

Express $\frac{7 \times 6}{4 \times 3}$ in terms of factorials.

View Detailed Solution ▼

Multiply by the missing numbers to create full factorial sequences: $\frac{(7 \times 6 \times 5!) \times 2!}{5! \times (4 \times 3 \times 2!)} = \mathbf{\frac{7! \times 2!}{5! \times 4!}}$

Equating Factorials

If $\frac{1}{(n-4)!} = \frac{20}{(n-2)!}$ , calculate $n$ .

View Detailed Solution ▼

Cross-multiply: $(n-2)! = 20(n-4)!$ .
Expand $(n-2)!$ to $(n-2)(n-3)(n-4)!$ .
$(n-2)(n-3)(n-4)! = 20(n-4)!$ .
$(n-2)(n-3) = 20$ . $n^2 - 5n + 6 = 20 \implies n^2 - 5n - 14 = 0$ $(n-7)(n+2) = 0$ Thus, $n = 7$ .

Basic Calculation

Find the value of the expression $\frac{6 \times 5 \times 4!}{4}$ .

View Detailed Solution ▼

$6 \times 5 = 30$ , and $4! = 24$ . $\frac{30 \times 24}{4} = 30 \times 6 = \mathbf{180}$

💡

The Countdown Analogy

Think of a factorial as a “Strict Countdown.”

If you are launching a rocket from T-minus 10 ( $10!$ ), you must count every single integer down to 1. If you decide to stop at T-minus 4 to take a break, you have completed the first part of the product ( $10 \times 9 \times 8 \times 7 \times 6 \times 5$ ) but you still have the “rest of the countdown” ( $4!$ ) waiting to be finished later.