Factorial notation

The governing concept is factorial notation. For a positive integer n, n! means the product of all positive integers from n down to 1. Therefore, 5! = 5 × 4 × 3 × 2 × 1. Calculating successively gives 5 × 4 = 20, 20 × 3 = 60, 60 × 2 = 120, and 120 × 1 = 120. Hence option B is correct. The value 60 is an intermediate product obtained before multiplying by 2, while 100 and 25 do not arise from the factorial definition. Factorial notation is used in counting arrangements and combinations, so preserving every factor down to 1 is essential; omitting a factor or stopping too early produces an incorrect result.

The factorial of a non-negative integer n is defined as n! = n × (n − 1) × ... × 2 × 1. The special case 0! is defined to be 1. This definition is necessary for the factorial rule n! = n × (n − 1)! to remain valid when n = 1, because 1! = 1 × 0!, giving 0! = 1. It also makes the combinations formula nCr = n!/[r!(n − r)!] work correctly when r = 0 or r = n. Therefore, option C is correct. Option A confuses the factorial symbol with the number itself, option B is incorrect because 0! is explicitly defined, and option D has no mathematical basis.

Here, 1! = 1 and 3! = 3 × 2 × 1 = 6. Therefore, 1! + 3! = 1 + 6 = 7. Option 6 is only the value of 3!; it does not include 1!. Exam tip: Evaluate each factorial first, then perform the indicated operation.

Since \(6!=6\times5\times4!\), \(\frac{6!}{4!}=\frac{6\times5\times4!}{4!}=6\times5=30\). Therefore, 30 is correct. The value 120 is \(5!\), not the value of this quotient. Exam tip: In factorial division, expand the larger factorial only up to the smaller factorial and cancel common terms.

The direct answer is B, 21. A factorial means multiplying all positive integers down to 1: \(7!=7\times6\times5\times4\times3\times2\times1\). In the expression, cancel \(5!\) from \(7!\): \(\frac{7!}{5!\,2!}=\frac{7\times6\times5!}{5!\times2}=\frac{7\times6}{2}=\frac{42}{2}=21\). Option A, 14, is wrong because it does not result from the required factorial cancellation. Option B, 21, is correct because it exactly equals the simplified expression. Option C, 35, is wrong; it may come from incomplete or incorrect multiplication and division. Option D, 42, is also wrong because 42 is the numerator \(7\times6\) before dividing by \(2!=2\). Remember: cancel the smaller factorial first, then divide by the remaining factorial.

By definition, \(n!=n\times(n-1)\times(n-2)\times\cdots\times1\). The remaining product is \((n-1)!\), so \(n!=n(n-1)!\). Option A uses addition, whereas a factorial is a product of consecutive positive integers. Exam tip: use \(n!=n(n-1)!\) to simplify factorial expressions.

Here, 4! = 4 × 3 × 2 × 1 = 24 and 2! = 2 × 1 = 2. Therefore, 4! + 2! = 24 + 2 = 26. The value 28 can result from an incorrect evaluation of 4!; evaluate each factorial first and then add. Exam tip: the symbol ! denotes the factorial of the number immediately before it.

Factorial notation represents a product of consecutive positive integers. Expand only the part needed for cancellation: 8! = 8 × 7!. Therefore, 8!/7! = (8 × 7!)/7! = 8, since 7! is nonzero and cancels from numerator and denominator. Option D is consequently correct. A common error is to multiply 8 by 7 and choose 56, as in option C; that would ignore the division by 7!. Option A gives the remaining factorial factor rather than the quotient, while option B results from adding 8 and 7 instead of simplifying the fraction. The efficient method is to use n! = n × (n − 1)! rather than calculate either factorial completely.

(9!=9\times8\times7\times6!), so (\frac{9!}{6!}=9\times8\times7). Divide by the factorial after the last required factor.

In \(n!\), all positive integers from \(n\) down to 1 are multiplied, so option A is correct. Option B represents addition, not factorial multiplication. Exam tip: identify factorial by the descending product ending in 1.

Since \(5!=5\times4\times3!\), \(\frac{5!}{3!}=\frac{5\times4\times3!}{3!}=5\times4=20\). Therefore, option C is correct. Option D is the value of \(5!\), but the expression also requires division by \(3!\). Exam tip: In factorial fractions, expand the larger factorial up to the smaller factorial and cancel the common factorial.

By definition, n! is the product of all positive integers from n down to 1. Thus, 3! = 3 × 2 × 1 = 6 and 2! = 2 × 1 = 2. Multiplying the two factorial values gives 3! × 2! = 6 × 2 = 12. Hence option A is correct. Option B may arise from adding 6 and 2 instead of multiplying them. Option C can result from an incorrect evaluation of one factorial, and option D is only the value of 3!, not the requested product. The important governing idea is that the multiplication sign outside the factorials must also be carried out after each factorial has been evaluated.

\(4!=4\times3\times2\times1=24\) and \(2!=2\times1=2\). Hence, \(\frac{4!}{2!}=\frac{24}{2}=12\). It can also be simplified as \(\frac{4\times3\times2!}{2!}=4\times3=12\). Option 8 would result from incorrectly using \(4\times2\), which is not the correct factorial expansion. Exam tip: In \(\frac{n!}{r!}\), expand the larger factorial up to \(r!\) and cancel the common factor.

Here, \(6!=720\) and \(3!=6\). Therefore, \(\frac{6!}{3!\,3!}=\frac{720}{6\times6}=\frac{720}{36}=20\). A value such as 18 may result from evaluating a factorial or multiplying the denominator incorrectly. Exam tip: write the values of smaller factorials first, then multiply and divide.

The factorial symbol means the product of all positive whole numbers from the given number down to 1. Thus \\(5!=5\\times4\\times3\\times2\\times1=120\\), while \\(4!=4\\times3\\times2\\times1=24\\). The expression asks for the difference between these two completed factorial values.

Substituting the values gives \\(5!-4!=120-24=96\\). A common mistake is to treat the expression as a factorial of the difference, but \\(5!-4!\\) means subtract two factorials, not calculate \\(1!\\). Another mistake is to stop at a partial product. Since the correct difference is 96, option A is the required answer.

By definition, \(0!=1\), \(1!=1\), and \(2!=2\times1=2\). Therefore, \(0!+1!+2!=1+1+2=4\). Option A results from omitting \(0!\), but \(0!\) is always 1. Exam tip: Write the values of small factorials before adding them.

Since \(10!=10\times9\times8!\), \(\frac{10!}{8!}=\frac{10\times9\times8!}{8!}=10\times9=90\). Option D is the value of \(6!\), not of the given expression. In exams, expand only up to the common factorial term and cancel it.

A factorial is the product of all positive integers from the given number down to 1. Therefore, \(4!=4\times3\times2\times1\), so option C is correct. Option B represents \(4^4\), not a factorial. Exam tip: Write \(n!\) as multiplication from \(n\) down to 1.

The product from (1) to (6) is (6!). In factorial multiplication, writing in reverse or forward order gives the same value.

Since \(9!\) is non-zero, a non-zero quantity divided by itself equals \(1\). Hence, \(\frac{9!}{9!}=1\). The value 9 is not the value of 9!, and 81 is \(9\times9\). Exam tip: When the same non-zero term occurs in the numerator and denominator, their ratio is always \(1\).

Using the factorial relation, \((n+1)!=(n+1)\cdot n!\). Therefore, \(\frac{(n+1)!}{n!}=\frac{(n+1)n!}{n!}=n+1\), so option B is correct. The distractor \(n\) may result from incorrectly treating \(n!\) as \(n\). Exam tip: expand the larger factorial in terms of the smaller one before cancelling.

Since \(7!=7\times6!\) and \(1!=1\), \(\frac{7!}{6!,1!}=\frac{7\times6!}{6!\times1}=7\). Therefore, the correct answer is 7. The number 6 is not the result; the factor \(6!\) cancels, leaving 7. Exam tip: In factorial fractions, expand the larger factorial up to the smaller factorial and cancel common factors.

Here, 3! = 3 × 2 × 1 = 6 and 4! = 4 × 3 × 2 × 1 = 24. Therefore, 3! + 4! = 6 + 24 = 30, so option B is correct. 24 is only the value of 4!, not the sum. Exam tip: In expressions involving factorials, evaluate each factorial first and then perform addition or subtraction.

A factorial is the product of all positive integers from the given number down to \(1\). Thus \(6!=6\times5\times4\times3\times2\times1=720\), while \(5!=5\times4\times3\times2\times1=120\). The expression asks for subtraction, not multiplication or division, so the two factorial values must be found and then subtracted in the stated order.

Substitute the factorial values: \(6!-5!=720-120\). Subtracting gives \(720-120=600\). Therefore the value of the expression is \(600\), which matches option A. A common error is to remember \(6!=720\) but forget to subtract \(5!=120\), or to confuse the expression with \(6\times5!\). The given answer is consistent.

\(2!=2\) and \(4!=4\times3\times2\times1=24\). Therefore, \(2!\times4!=2\times24=48\). The value 36 is obtained from \(3!\times3!\), not from the given expression. In exams, evaluate each factorial separately before multiplying.