5! is computed as 5 × 4 × 3 × 2 × 1, which equals 120. This demonstrates the basic principle of factorial operations.

n! denotes the product of all positive integers up to n. This definition is fundamental for understanding factorials in mathematics.

By definition, 0! is equal to 1. This convention is used to simplify formulas in combinatorics and other areas of mathematics.

3! is calculated as 3 × 2 × 1, which equals 6. This is a straightforward application of the factorial definition.

4! means multiplying all positive integers from 1 to 4, which is 4 × 3 × 2 × 1. This option correctly represents the factorial operation.

Since 6! equals 6 multiplied by 5!, the expression can be written as 6 × 5!. This showcases the recursive nature of factorials.

Dividing 7! by 5! cancels out the common factorial terms, leaving 7 × 6, which equals 42. This problem emphasizes how factorial division works.

Factorials grow at an extremely rapid rate because they involve the product of an increasing series of numbers. Recognizing this property is essential for understanding factorial behavior.

The number of trailing zeros in a factorial is determined by the number of times 10 is a factor, which depends on the number of pairs of 2 and 5. In 10!, there are two factors of 5, resulting in 2 trailing zeros.

The recursive definition of factorial is given by n! = n × (n - 1)!. This is a key concept that underpins many properties and computations involving factorials.

Since 8! can be expressed as 8 × 7!, dividing 40320 by 8 yields 5040. This relationship between consecutive factorials is useful for quick computations.

Since (n+1)! is equal to (n+1) × n!, the n! cancel, leaving n + 1. This is a standard simplification when working with factorial expressions.

4! equals 24 and 3! equals 6; adding these gives 30. This problem reinforces the arithmetic operations involving factorials.

Factorials increase as the integer increases, making 6! the largest among the options. This question highlights the rapid growth of factorial values.

Since 5! equals 120, it follows that x must be 5. Recognizing common factorial values helps in solving such equations quickly.

9! divided by 7! simplifies to 9 × 8, which equals 72. This requires cancelling common factorial terms to simplify the expression.

Since 9! is equal to 362880, the value of n must be 9. Familiarity with common factorial values is essential for this type of problem.

Substituting n = 4 gives (6!)/(4!), which simplifies to 6 × 5 = 30. This problem tests the ability to cancel and simplify factorial expressions.

Factoring out n! from (n+1)! - n! gives n![(n+1) - 1] which simplifies to n × n!. This demonstrates proper factoring in factorial expressions.

Calculating 5! gives 120 and 2! gives 2, so 120 × 2 equals 240; subtracting 3! (which is 6) results in 234. This problem combines multiple factorial computations.