Factorial notation

Since \(11!=11\times10\times9!\), \(\frac{11!}{9!}=\frac{11\times10\times9!}{9!}=11\times10=110\). Therefore, the correct answer is 110. Note that 121 is \(11\times11\), whereas the required consecutive factors are 11 and 10. Exam tip: In factorial division, expand the larger factorial only up to the smaller factorial and cancel common factors.

In factorial notation, \(n!\) is the product of all positive integers from \(1\) to \(n\). Since \(7!=7\times6\times5\times4\times3\times2\times1=5040\), \(n=7\). The closest distractor, \(8\), is incorrect because \(8!=8\times5040=40320\). Exam tip: remember common values such as \(5!=120\), \(6!=720\), and \(7!=5040\).

Since \((n+2)!=(n+2)(n+1)!\), \(\frac{(n+2)!}{(n+1)!}=n+2\). Adding 1 gives \(n+2+1=n+3\). Option \(n+2\) misses the \(+1\) outside the fraction. Exam tip: In factorial fractions, expand the larger factorial up to the smaller factorial and cancel common factors.

Here, \(2!=2\), \(3!=3\times2\times1=6\), and \(4!=4\times3\times2\times1=24\). Therefore, \(2!\times3!\times4!=2\times6\times24=288\). Option D, \(48\), is only \(2!\times4!\) and misses \(3!\). Exam tip: Evaluate each factorial separately before multiplying.

Since \(6!=6\times5\times4!\), \(\frac{6!}{4!}=6\times5=30\). Therefore, the ratio is \(30:1\). The ratio \(20:1\) would result from incorrectly using only \(5\times4\). In exams, expand the larger factorial up to the smaller factorial and cancel the common part.

\(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\), so \(6!\) is the correct option. \(5! = 120\), making it a close but incorrect distractor. Exam tip: if you remember \(5! = 120\), multiply it by 6 to find \(6! = 720\) quickly.

First, evaluate the bracket: \(6-3=3\). Thus, \(3!=3\times2\times1=6\) and \(2!=2\times1=2\). Therefore, \(3!\times2!=6\times2=12\). Option 10 is incorrect because \(3!=6\), not 3. Exam tip: In factorial expressions, simplify brackets first and then evaluate each factorial.

By the factorial definition, \(n!=n\times(n-1)!\). Therefore, \(14!=14\times13!\), so the correct answer is \(14!\). The expression \(15!\) would require \(15\times14!\), so it is not correct. Exam tip: multiplying a factorial by its next integer gives the factorial of that integer.

Since \(5!=5\times4!\) and \(1!=1\), \(\frac{5!}{1!,4!}=\frac{5\times4!}{1\times4!}=5\). The value 10 is \(\binom{5}{2}\), but the denominator here is \(1!\,4!\); hence the expression equals \(\binom{5}{1}=5\). Exam tip: In factorial fractions, cancel common factorial terms first.

\(4!=24\), \(2!=2\), and \(3!=6\). Hence, \(4!\times2!-3!=24\times2-6=48-6=42\). Therefore, option C is correct. \(48\) is a close distractor obtained by forgetting to subtract \(3!\). Exam tip: evaluate factorials first, then apply multiplication and subtraction in order.

(9!=9\times8!), so (8!=\frac{9!}{9}). A smaller factorial can also be found from a larger factorial.

The factorial symbol means multiplying all positive integers down to 1. In a quotient of factorials, the smaller factorial can be cancelled from the larger one. This leaves only the factors that occur in the larger factorial but not in the smaller factorial. The required option is C, because the remaining product is 132.

Write \(12!\) as \(12\times11\times10!\). Then \(\frac{12!}{10!}=\frac{12\times11\times10!}{10!}=12\times11=132\). The \(10!\) factors cancel, not the factors 12 and 11. Therefore option C is correct. Options 110 and 120 do not use the correct two remaining factors, while 144 is an incorrect product.

By definition, \(0! = 1\) and \(1! = 1\). Therefore, \(p+q = 0!+1! = 1+1 = 2\). Option 1 is the value of either factorial individually, not their sum. Exam tip: Always remember that \(0! = 1\).

\(7!\div5!=\frac{7\times6\times5!}{5!}=7\times6=42\). Also, \(2!=2\times1=2\). Therefore, the required difference is \(42-2=40\). Option 42 is only the value of \(7!\div5!\), not the difference. Exam tip: In factorial division, cancel the common factorial term first to simplify the calculation.

The direct answer is B, 1. By definition, \(5!=5\times4\times3\times2\times1\). The factors decrease by one each time and stop at 1; the final factor written in this product is therefore 1. Option A, 0, is wrong because zero is not included in the usual factorial product, and including it would make the product zero. Option B, 1, is correct because every positive-integer factorial ends with multiplication by 1. Option C, 2, is wrong because 2 is the second-to-last factor in \(5!\), not the last one. Option D, 5, is wrong because 5 is the first, largest factor. A factorial is not just the number itself: it is the product of all positive integers from that number down to 1. Memory cue: factorial always walks downward and finishes at 1; separately, \(0!=1\) is a special definition.

Since \(6!=6\times5\times4\times3!\), \(\frac{6!}{3!}=6\times5\times4=120\). Now \(120\div5=24\), so the correct answer is 24. A value such as 30 can result from cancelling factors incorrectly. Exam tip: In a factorial fraction, first expand only up to the common factorial and cancel it before calculating.

Expanding the larger factorial gives \((n+3)!=(n+3)(n+2)(n+1)!\). Dividing by \((n+1)!\) therefore leaves \((n+3)(n+2)\). Option C incorrectly includes the extra factor \((n+1)\). Exam tip: In factorial division, expand the larger factorial only up to the smaller factorial and cancel the common part.

A factorial is the product of all positive integers from the given number down to 1. Thus, the expression must be evaluated as a difference of two factorials: first calculate 6!, then calculate 4!, and finally subtract the smaller value from the larger one. It is important not to read 6!-4! as an ordinary difference inside a factorial; the factorial signs apply separately to 6 and 4.

We have \(6!=6\times5\times4\times3\times2\times1=720\), while \(4!=4\times3\times2\times1=24\). Therefore, \(6!-4!=720-24=696\). Hence option A is correct. The other numerical choices result from an arithmetic error or from mishandling one of the factorials.

A factorial is the product of all positive integers from the given number down to 1. Hence, \(9!=9\times8\times7\times\cdots\times1\). Option B uses addition and option D uses subtraction, while option C incorrectly repeats 9. Exam tip: Remember \(n! = n\times(n-1)\times\cdots\times1\).

\(4!=4\times3\times2\times1=24\), and \(1!=1\). Hence, \(\frac{4!+1!}{5}=\frac{24+1}{5}=\frac{25}{5}=5\). Therefore, 5 is the correct option. Option 4 may result from incorrectly handling the addition of \(1!\). Exam tip: evaluate each factorial first, then perform addition or division.

Using factorials, 2! = 2 and 4! = 4 × 3 × 2 × 1 = 24. Therefore, m + n = 2 + 24 = 26, so option C is correct. Note that 24 is only the value of n, not of m + n. Exam tip: Evaluate each factorial first, then substitute the values into the given expression.

By the definition of factorial, \(15!=15\times14\times13\times\cdots\times1\), while \(14!=14\times13\times\cdots\times1\). Hence, \(15!=15\times14!\). Option C still contains \(15!\) itself, so it does not correctly express it in terms of \(14!\). Exam tip: remember the general identity \(n!=n(n-1)!\).

Here, \(\frac{10!}{8!,2!}=\frac{10!}{8!\times2!}\). Cancelling \(8!\) from \(10!\) gives \(\frac{10\times9}{2\times1}=45\). Therefore, 45 is the correct answer. Values such as 40 or 50 can result from an error in multiplication or in dividing by \(2!\). Exam tip: In factorial fractions, expand the larger factorial only as far as needed and cancel the common factorial first.

Using factorials, \(3!=3\times2\times1=6\), \(4!=4\times3\times2\times1=24\), and \(0!=1\). Therefore, \(3!+4!\times0!=6+24\times1=30\). The value 24 is only \(4!\times0!\); it incorrectly omits adding \(3!\). Exam tip: evaluate factorials first, then multiplication, and finally addition.

Since \(8!=8\times7!\), \(\frac{8!}{8}=\frac{8\times7!}{8}=7!\). Therefore, \(7!\) is correct. Choosing \(8!\) is incorrect because dividing by 8 removes the factor 8. Exam tip: use \(n!=n\times(n-1)!\) to simplify such expressions quickly.