Factorial notation

Using the factorial rule, \(10!=10\times9\times8!\), because \(10!=10\times9\times8\times7\times\cdots\times1\) and \(8!=8\times7\times\cdots\times1\). Therefore, \(10\times9\times8!\) is correct. In \(10\times8!\), the factor \(9\) is missing. Exam tip: To express a larger factorial using a smaller one, multiply by the consecutive descending integers up to that smaller factorial.

Since \((n+1)!=(n+1)\cdot n\cdot (n-1)!\), \(\frac{(n+1)!}{(n-1)!}=\frac{(n+1)n(n-1)!}{(n-1)!}=n(n+1)\). Option D incorrectly uses \(n-1\) in place of \(n\). Exam tip: In factorial fractions, expand the larger factorial up to the smaller factorial before cancelling.

By factorial notation, \(2!=2\times1=2\) and \(5!=5\times4\times3\times2\times1=120\). Therefore, \(2!\times5!=2\times120=240\), so option C is correct. Option A is only the value of \(5!\); it does not include multiplication by \(2!\). Exam tip: Evaluate each factorial first, then perform the required operation.

Since \(6!=6\times5\times4!\), \(\frac{6!}{4!}=6\times5=30\). Also, by definition, \(0!=1\). Therefore, \(\frac{6!}{4!}+0!=30+1=31\). Option 30 results if \(0!\) is not added. Exam tip: In factorial fractions, cancel common factorial terms first to simplify quickly.

Since \(4!=4\times3!\), \(4!:3!=4\times3!:3!=4:1\). Hence, the correct ratio is \(4:1\). The ratio \(3:1\) would arise from comparing \(3!\) with \(2!\). Exam tip: use \(n!=n\times(n-1)!\) to simplify factorial ratios quickly.

Using factorials, \(5!=5\times4\times3\times2\times1=120\) and \(3!=3\times2\times1=6\). Therefore, \(5!-3!=120-6=114\). The value 120 is only \(5!\); it does not include subtracting \(3!\). Exam tip: Evaluate each factorial first, then perform the indicated operation.

\(3!=3\times2\times1=6\), \(0!=1\), and \(1!=1\). Therefore, \(\frac{3!+0!}{1!}=\frac{6+1}{1}=7\). Option 6 would result from incorrectly taking \(0!\) as 0. Exam tip: always remember that \(0!=1\) and \(1!=1\).

For factorials, \(n!=n\times(n-1)\times\cdots\times1\). Thus, \(4!=4\times3\times2\times1=24\), so \(4!\) is correct. In comparison, \(3!=6\) and \(5!=120\), so neither equals 24. Exam tip: memorise common small factorial values such as \(3!=6\), \(4!=24\), and \(5!=120\).

Using the factorial relation, \(10!=10\times 9!\). Dividing both sides by 10 gives \(9!=\frac{10!}{10}\). \(\frac{10!}{9}\) is incorrect because to obtain 9! from 10!, we divide by the extra factor 10. Exam tip: Use \(n!=n\times (n-1)!\) to simplify factorial expressions quickly.

(\frac{8!}{4!,4!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70). Cancel factorials to make the numbers smaller.

By definition, \(6!=6\times5\times4\times3\times2\times1\). The product contains 6, 5, 4, 3, 2, and 1, so there are 6 factors in total. Choosing 5 would incorrectly omit the final factor, 1. Exam tip: \(n!\) contains \(n\) factors, from 1 to \(n\).

Since \(7! = 7\times6\times5!\), \(\frac{7!}{5!}=7\times6=42\). Also, \(1!=1\), so \(42+1=43\). Option 42 results from missing the addition of \(1!\). Exam tip: In a factorial quotient, cancel the common factorial term and multiply the remaining factors.

Here, \(3!=6\) and \(4!=24\). Therefore, \(3!\times4!-4!=6\times24-24=144-24=120\). Hence, option C is correct. \(144\) results if the final \(4!\) is not subtracted. Exam tip: In expressions involving multiplication and subtraction, evaluate factorials first, then multiply, and finally subtract.

Here, \(5!=5\times4\times3\times2\times1=120\) and \(4!=4\times3\times2\times1=24\). Therefore, \(x-y=120-24=96\). Option 120 is only the value of \(5!\); it does not subtract \(4!\). Exam tip: In factorial questions, evaluate each factorial first and then perform the required operation.

Evaluate the bracket first: \(4-2=2\). Thus, the expression becomes \(2!\). Since \(2!=2\times1=2\), the correct answer is 2. Directly taking the factorial of 4 would give \(4!=24\), which is not the given expression. Exam tip: always simplify the expression inside brackets before evaluating a factorial.

By the definition of factorial, \(5!=5\times4\times3\times2\times1\), while \(3!=3\times2\times1\). Hence, \(5!=5\times4\times3!\). In \(5\times3!\), the factor \(4\) is missing. In exams, expand the larger factorial only up to the given smaller factorial.

Since \(11!=11\times10!\), \(\frac{11!}{10!}=\frac{11\times10!}{10!}=11\). The value 110 would result from multiplying 11 by 10, but here the entire \(10!\) is cancelled. Exam tip: use \(\frac{n!}{(n-1)!}=n\) directly.

Here, 6! = 6×5×4×3×2×1 = 720. Also, 0! = 1 and 1! = 1. Therefore, 6!+0!-1! = 720+1-1 = 720, so option C is correct. The value would be 721 if 1! were not subtracted. Exam tip: always remember that 0! = 1.

Using factorial expansion, \((n+4)!=(n+4)(n+3)!\). Therefore, \(\frac{(n+4)!}{(n+3)!}=\frac{(n+4)(n+3)!}{(n+3)!}=n+4\). Although \(n+3\) is a nearby term, it belongs to the factorial in the denominator and cancels out; the extra factor left is \(n+4\). Exam tip: apply \(\frac{(k+1)!}{k!}=k+1\) directly.

The phrase asks for the sum of two quantities, so interpret the expression as (4! ÷ 2!) + 3!. Evaluate the factorials first: 4! = 4 × 3 × 2 × 1 = 24, 2! = 2, and 3! = 3 × 2 × 1 = 6. Therefore, 4! ÷ 2! = 24 ÷ 2 = 12, and the required sum is 12 + 6 = 18. Hence option B is correct. Option A can result from an incorrect factorial or addition calculation. Option C may arise if the quotient is mistakenly treated as 14. Option D is only 4! and does not include the required 3!. Separating factorial evaluation, division, and addition avoids ambiguity and follows the order of operations.

By the factorial identity, \(n!=n\times(n-1)!\). Hence, \(9!=9\times8!\), so \(9!\) is the correct option. \(10!\) would require \(10\times9!\), so it is not correct. Exam tip: Rewrite \(n!\) as \(n\times(n-1)!\) to solve such questions quickly.

Since \(5!=120\), \(2!=2\), and \(3!=6\), \(\frac{5!}{2!\,3!}=\frac{120}{2\times6}=10\). Therefore, the correct answer is 10. The value 12 may result from not multiplying the factorials in the denominator correctly. Exam tip: evaluate the smaller factorials first, then divide.

By factorial notation, \(2!=2\times1=2\) and \(5!=5\times4\times3\times2\times1=120\). Therefore, \(2!+5!=2+120=122\), so option C is correct. \(120\) is a close distractor because it results from forgetting to add \(2!\). Exam tip: evaluate each factorial separately before carrying out the indicated operation.

By definition, \(7!=7\times6\times5\times4\times3\times2\times1\). Since \(5!=5\times4\times3\times2\times1\), we get \(7!=7\times6\times5!\). In \(7\times5!\), the factor \(6\) is missing. Exam tip: When expressing \(n!\) in terms of \((n-r)!\), include every consecutive factor in between.

(12!=12\times11\times10!), so the given form is (12!). Recognize factorial expansion down to a smaller factorial.