Factorial notation

Since \(8!=8\times7\times6\times5!\), \(\frac{8!}{5!}=8\times7\times6=336\). The value 120 is \(5!\), but \(5!\) cancels in this expression. Exam tip: In a factorial fraction, expand the larger factorial only up to the smaller factorial and cancel common terms.

Given n=3, we get n+2=3+2=5. Therefore, (n+2)!=5!=5×4×3×2×1=120. Hence, 120 is the correct answer. The value 24 is equal to 4!, so it is not correct here. Exam tip: Evaluate the expression inside the brackets before applying the factorial.

Since \((n+2)!=(n+2)(n+1)n!\), we get \(\frac{(n+2)!}{n!}=\frac{(n+2)(n+1)n!}{n!}=(n+2)(n+1)\). The option \(n(n+1)\) incorrectly omits the factor \(n+2\). Exam tip: expand a factorial only up to the common factorial term, then cancel it.

(5!=120) and (4!=24), so (5!) is greater. The factorial of a larger positive integer is generally larger.

Since \(4!=4\times3!\), \(\frac{4!+3!}{3!}=\frac{4\times3!+3!}{3!}=\frac{5\times3!}{3!}=5\). Option 4 would result if the \(+3!\) term in the numerator were incorrectly ignored. Exam tip: Factor out the common factorial term first to simplify such expressions quickly.

Factorial notation is defined for a positive integer n by n! = n × (n − 1) × (n − 2) × ... × 2 × 1. Applying this definition gives 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1. The factors decrease by one at each step, so the factor immediately after 7 is 6. Therefore, option C is correct. Option A, 8, would appear at the beginning of 8!, not 7!. Option B incorrectly repeats the first factor, while option D is the last factor in the expansion, not the factor immediately following 7. The question tests direct recognition of the descending consecutive-product pattern used in factorial notation.

\(3!=3\times2\times1=6\), and by definition \(0!=1\). Therefore, \(3!\times0!=6\times1=6\), so option C is correct. It is incorrect to treat \(0!\) as 0; \(0!=1\) keeps permutation and combination formulas consistent. Exam tip: whenever you see \(0!\), substitute 1.

The expression is a factorial quotient. Use 6! = 6 × 5 × 4!, while 2! = 2. Then 6!/(2! × 4!) = (6 × 5 × 4!)/(2 × 4!) = (6 × 5)/2 = 30/2 = 15. Thus option C is correct. This expression is also the combination value 6C2, which counts selections of two objects from six, and it equals 15. Option A, 10, can come from using an incorrect cancellation; option B misses the division by 2; and option D results from failing to divide by the remaining 2. The safest method is to cancel the common 4! before performing the arithmetic.

The recursive factorial rule is \(n! = n\times(n-1)!\); for example, \(4!=4\times3!\). Option B incorrectly uses addition, and \(0!=1\), not 0. Exam tip: always remember the special value of 0!

(6!=6\times5\times4!), so the given product is (6!). A factorial can be identified from the middle expansion too.

\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Therefore, \(\frac{5!}{5}=\frac{120}{5}=24\). Hence, 24 is the correct answer. Choosing 20 results from not multiplying all the factors of \(5!\) correctly. Exam tip: use \(n! = n \times (n-1)!\) to simplify factorial fractions quickly.

The governing concept is factorial notation. For a positive integer n, n! means the product n × (n − 1) × ... × 2 × 1. The factorial signs in this expression apply separately to 4 and 3, so calculate each one before subtracting. We get 4! = 4 × 3 × 2 × 1 = 24 and 3! = 3 × 2 × 1 = 6. Therefore, 4! − 3! = 24 − 6 = 18, so option A is correct. The subtraction is not included inside either factorial. Treating the exclamation mark as an ordinary multiplication symbol or applying one factorial to the entire difference would produce an invalid method. The other options do not equal the calculated difference.

The correct answer is C: x=24. The symbol 4! is read as “four factorial.” Factorial means multiplying the positive integers from 4 down to 1: 4!=4×3×2×1=24. Since the question states x=4!, replace the factorial by its value and obtain x=24. Option A, 16, is incorrect because it is 4×4, not factorial. Option B, 20, does not follow from the factorial calculation. Option C is correct. Option D, 28, is also not the product 4×3×2×1. Remember that 0!=1, while for a positive integer n, n!=n(n-1)(n-2)…1. Exam cue: expand the factorial completely before selecting an option.

Since \(3!=3\times2!\), \(\frac{3!}{2!}=\frac{3\times2!}{2!}=3\). Therefore, the correct answer is 3. The value 6 is only the value of \(3!\); it ignores division by \(2!\). Exam tip: Remember that \(\frac{n!}{(n-1)!}=n\).

The governing concept is the factorial identity n! = n × (n − 1)!, together with the definition 3! = 3 × 2 × 1. Substitute this expansion into the given expression: 5 × 4 × 3! = 5 × 4 × (3 × 2 × 1) = 5 × 4 × 3 × 2 × 1. This product is exactly the expansion of 5!, so the correct answer is option C, 5!. Option A is only 3! and leaves out the factors 5 and 4. Option B is 4! and omits the factor 5. Option D would require an additional factor 6, which is not present. Recognizing the consecutive factorial pattern is quicker and safer than multiplying all factors separately.

Factorial notation means multiplying every positive integer from the stated number down to 1. Thus, 6! = 6 × 5 × 4 × 3 × 2 × 1. Computing in convenient groups gives 6 × 5 = 30, 4 × 3 = 12, and 2 × 1 = 2; therefore 30 × 12 × 2 = 720. Hence option C is correct. A useful independent check is the recurrence 6! = 6 × 5!, and since 5! = 120, we get 6 × 120 = 720. Option A is only 5!, while options B and D result from incomplete or incorrect multiplication. The definition and recurrence both confirm the same answer.

Since \(7!=7\times6!\), \(\frac{7!}{7}=\frac{7\times6!}{7}=6!\). Getting \(5!\) would require dividing \(6!\) by 6 as well, so it is not correct. Exam tip: use \(n!=n\times(n-1)!\) to simplify factorial expressions.

\(4!=4\times3\times2\times1=24\), and by definition \(0!=1\). Therefore, \(\frac{4!}{0!}=\frac{24}{1}=24\). Option 1 is incorrect because it is the value of \(0!\) only, not of the complete quotient. Exam tip: always remember that \(0!=1\).

Given \(n!=120\), the factorial of \(n\) is 120. Since \(5!=5\times4\times3\times2\times1=120\), \(n=5\). In contrast, \(4!=24\), so option \(4\) is not correct. Exam tip: Memorising factorial values from \(3!\) to \(6!\) helps solve such questions quickly.

First evaluate the expression inside the brackets: \(3+1=4\). Then \(4!=4\times3\times2\times1=24\). Therefore, the correct answer is 24. Choosing 6 would mean using \(3!\) and ignoring the addition inside the brackets. Exam tip: always simplify the expression inside brackets before applying factorial.

Here, \(3! = 3 \times 2 \times 1 = 6\), and by definition \(0! = 1\). Therefore, \(3!+0! = 6+1 = 7\). Option 6 is only the value of \(3!\); it does not include \(0!\). Exam tip: Always remember that \(0! = 1\).

By the definition of factorial, \(8!=8\times 7\times 6\times\cdots\times 1\). Since \(7!=7\times 6\times\cdots\times 1\), we get \(8!=8\times 7!\). The expression \(8\times 6!\) misses the factor 7, so it is not equal to \(8!\). Exam tip: use the identity \(n!=n\times (n-1)!\).

Since \(5!=5\times4!\), we get \(\frac{5!}{4!}=5\). Therefore, the given expression is \(5+1=6\). It is not necessary to evaluate both factorials completely; \(4!\) cancels directly. Exam tip: For consecutive factorials, write the larger factorial in terms of the smaller one before simplifying.

Here, \(2!=2\) and \(3!=6\). Therefore, \(\frac{2!+3!}{2!}=\frac{2+6}{2}=\frac{8}{2}=4\). Option 5 may result from incorrectly taking \(3!\) as 8. Exam tip: evaluate each factorial first, then perform the indicated operations.

(7!=7\times6\times5\times4\times3\times2\times1), which has (7) factors. In (n!), factors run from (n) down to (1).