Factorial notation

Writing \(10!=10\times9!\), the numerator becomes \(10\times9!+9!=9!(10+1)=11\times9!\). Hence, \(\frac{11\times9!}{8!}=11\times9=99\). Therefore, 99 is the correct answer. A value such as 108 may result from an incorrect addition while taking the common factorial. Exam tip: In factorial fractions, first express numerator terms using the same smaller factorial.

Since \(14!=14\times13!\), we get \(14!+13!=14\times13!+13!=13!(14+1)=15\times13!\). As \(13!=13\times12!\), \(\frac{15\times13!}{12!}=15\times13=195\). Hence, option B is correct. \(210\) may result from incorrectly treating the expression as \(14\times15\). Exam tip: for sums of factorials, first factor out the smaller common factorial.

Since \(n!=n(n-1)(n-2)!\), \(\frac{n!}{(n-2)!}=n(n-1)\). Thus, \(n(n-1)=132\). As \(12\times11=132\), \(n=12\). For the closest distractor, \(n=11\) gives \(11\times10=110\), not 132. Exam tip: Expand the larger factorial up to the smaller factorial and cancel common factors.

Write \((n+4)!=(n+4)(n+3)(n+2)!\). Cancelling \((n+2)!\) gives \(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\). Thus, \((n+4)(n+3)=156\). Since \(12\times13=156\), we have \(n+3=12\), so \(n=9\). For example, if \(n=10\), the value would be \(14\times13=182\), so it is not correct. Exam tip: In a factorial quotient, expand the larger factorial only up to the smaller factorial and then cancel common terms.

(72=2^3\times3^2) and (6!) contains all these factors. In such questions write prime factors first.

Since \(125=5^3\), \(n!\) must contain at least three factors of \(5\). In \(14!\), only \(5\) and \(10\) contribute factors of \(5\), so it has only two such factors and is not divisible by \(125\). In \(15!\), the terms \(5,10,15\) provide three factors of \(5\); hence \(15!\) is divisible by \(5^3=125\). Exam tip: for divisibility by \(p^k\), count the exponent of the prime \(p\) in \(n!\).

The number of zeros is (\left\lfloor\frac{50}{5}\right\rfloor+\left\lfloor\frac{50}{25}\right\rfloor=10+2=12). Counting powers of (5) is the fastest method.

The exponent is (\left\lfloor\frac{100}{5}\right\rfloor+\left\lfloor\frac{100}{25}\right\rfloor=20+4=24). For highest power add the quotients.

The exponent is (\left\lfloor\frac{80}{7}\right\rfloor+\left\lfloor\frac{80}{49}\right\rfloor=11+1=12). Add only up to powers of the prime.

We have \(25!=25\times24\times23!\). Therefore, \(\frac{25!}{23!}=\frac{25\times24\times23!}{23!}=25\times24=600\). Hence, option B is correct. In contrast, \(625=25\times25\), whereas the required consecutive factors are \(25\) and \(24\). Exam tip: In \(\frac{n!}{(n-r)!}\), multiply only the consecutive factors that remain after cancellation.

Since \(16!=16\times15\times14!\), \(\frac{16!}{14!\times8}=\frac{16\times15\times14!}{14!\times8}=\frac{16\times15}{8}=2\times15=30\). Hence, the correct answer is 30. Option 32 is a close distractor, but it does not follow from the correct simplification. Exam tip: In a factorial fraction, expand the larger factorial only up to the smaller factorial and cancel common factors first.

Since \((n+3)!=(n+3)(n+2)(n+1)!\), \(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\). Similarly, \(\frac{(n+2)!}{n!}=(n+2)(n+1)\). Hence, the expression becomes \((n+2)[(n+3)-(n+1)]=2(n+2)\). The option \(n+2\) incorrectly omits the difference factor 2. Exam tip: In factorial fractions, expand only until the denominator factorial and then cancel common factors.

The numerator becomes (n!(n+2)) and division gives (n(n+2)=150). Check carefully because (10\times12=120) is not enough.

By definition, \((n+1)!=(n+1)\cdot n!\). Indeed, \((n+1)!=(n+1)\cdot n\cdot(n-1)\cdots1\), while \(n!=n\cdot(n-1)\cdots1\). Therefore, option A is correct. Option B wrongly uses addition; factorials are related through multiplication. Exam tip: to simplify factorial expressions, separate the greatest factor from the larger factorial.

The value is ((7\times6)\times\frac{1}{6\times5}= \frac{7}{5}) not an integer. Since options do not match, read the expression carefully.

The first part is (8\times7) and the second is (7\times6). The ratio is (\frac{56}{42}=\frac{4}{3}), so it will not match these options.

The correct answer is 6 because \(6!=6\times5\times4\times3\times2\times1=720\). In contrast, \(5!=120\) and \(7!=5040\), so neither equals 720. Exam tip: Memorising factorial values from \(1!\) to \(7!\) helps solve such questions quickly.

Since \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\), we have \((n-1)! = 7!\). Therefore, \(n-1=7\), which gives \(n=8\). Option 7 is the value of \(n-1\), not of \(n\). Exam tip: First identify the known factorial equal to the given number, then solve for the variable.

Since \(0!=1\), \(1!=1\), \(2!=2\), and \(3!=6\), the expression becomes \(1+1+\frac{1}{2}+\frac{1}{6}\). Using \(\frac{1}{2}=\frac{3}{6}\), the sum is \(2+\frac{3}{6}+\frac{1}{6}=2+\frac{4}{6}=\frac{8}{3}\). The distractor \(\frac{5}{2}\) may result from leaving out the \(\frac{1}{3!}\) term. Exam tip: always write \(0!=1\) explicitly in factorial calculations.

Here, \(0!=1\), \(1!=1\), \(2!=2\), and \(3!=6\). Thus, the numerator is \(1+1+2+6=10\), while the denominator is \(4!=24\). Hence, the value is \(\frac{10}{24}=\frac{5}{12}\). The distractor \(\frac{3}{8}=\frac{9}{24}\) can arise from adding the numerator incorrectly. Exam tip: always include the fact that \(0!=1\) when evaluating factorial expressions.

Since \(4!=24\), we get \(m=4\). Similarly, \(6!=720\), so \(n=6\). Hence, \((n-m)!=(6-4)!=2!=2\). The value 6 in option C is \(3!\), not \(2!\). Exam tip: Remember small factorial values such as \(4!=24\), \(5!=120\), and \(6!=720\).

After cancelling \((n-4)!\), we get \(\frac{n!}{(n-4)!}=n(n-1)(n-2)(n-3)\). Since \(9\times8\times7\times6=3024\), \(n=9\). For the closest distractor, \(n=10\) gives \(10\times9\times8\times7=5040\), not 3024. Exam tip: rewrite such factorial ratios as products of consecutive decreasing integers.

\(\frac{17!}{15!}=17\times16\) because \(15!\) cancels. Similarly, \(\frac{16!}{14!}=16\times15\). Therefore, the value is \(17\times16-16\times15=16(17-15)=32\). Values such as 30 or 34 can result from subtracting the factors incorrectly. Exam tip: In factorial fractions, expand only up to the smaller factorial and cancel the common part.

Since \(19! = 19\times18\times17!\), \(\frac{19!}{17!}=19\times18\). Similarly, \(18!=18\times17\times16!\), so \(\frac{18!}{16!}=18\times17\). Therefore, the value is \(19\times18-18\times17=18(19-17)=36\). The nearby value 34 may result from incorrect cancellation of factorial terms. Exam tip: Expand a factorial only up to the smaller factorial before cancelling common factors.

For (8!), the exponent of (3) is only (\left\lfloor\frac{8}{3}\right\rfloor=2), so it is not correct. Checking properly gives exponent (4) for (9!).