Factorial notation

Write \((n+4)!=(n+4)(n+3)!\). The numerator then becomes \((n+3)![(n+4)+1]=(n+3)!(n+5)\). Dividing by \((n+2)!\) gives \((n+3)(n+5)=143\). Since \(143=11\times13\), we have \(n+3=11\), hence \(n=8\). For option 7, the value becomes \(10\times12=120\), not 143. Exam tip: In factorial fractions, express the larger factorial as a multiple of the smaller factorial before simplifying.

Using \((n+5)!=(n+5)(n+4)!\), the numerator becomes \((n+4)!\big[(n+5)-1\big]=(n+4)!(n+4)\). Also, \((n+4)!=(n+4)(n+3)!\). Therefore, \(\frac{(n+4)!(n+4)}{(n+3)!}=(n+4)^2\), so option A is correct. \((n+4)(n+5)\) would arise from incorrectly multiplying factors before carrying out the subtraction. Exam tip: in a factorial subtraction, first factor out the smaller common factorial.

Using \((n+5)!=(n+5)(n+4)!\), the numerator becomes \((n+4)![(n+5)-1]=(n+4)(n+4)!\). Therefore,
\[
\frac{(n+5)!-(n+4)!}{(n+3)!}
=\frac{(n+4)(n+4)!}{(n+3)!}
=(n+4)^2.
\]
Thus \((n+4)^2=169=13^2\). For all the factorials to be defined, \(n\geq -3\), so \(n+4\) is positive and \(n+4=13\). Hence \(n=9\). Option 8 gives \(12^2=144\), not 169. Exam tip: In factorial expressions, rewrite the larger factorial in terms of the smaller one before cancelling common factors.

( \frac{n!}{(n-3)!}=n(n-1)(n-2) ) and ( \frac{n!}{(n+3)!}=\frac{1}{(n+1)(n+2)(n+3)} ). Multiply both parts.

Putting (n=6) in the simplified form gives ( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} ). Cancel while checking options.

The exponent is ( \left\lfloor\frac{50}{2}\right\rfloor+\left\lfloor\frac{50}{4}\right\rfloor+\left\lfloor\frac{50}{8}\right\rfloor+\left\lfloor\frac{50}{16}\right\rfloor+\left\lfloor\frac{50}{32}\right\rfloor=47 ). Adding all quotients is necessary.

The number of zeros is ( \left\lfloor\frac{60}{5}\right\rfloor+\left\lfloor\frac{60}{25}\right\rfloor=14 ). Always add the extra contribution of (25).

Since \(23!=23\times22\times21!\), the first term is \(\frac{23\times22}{2!}=253\). Similarly, \(22!=22\times21\times20!\), so the second term is \(\frac{22\times21}{2!}=231\). Hence, the value is \(253+231=484\). Option 474 is incorrect and can result from an error in multiplication or addition. Exam tip: In factorial fractions, cancel the common factorial first and then multiply the remaining factors.

The first term is \(\frac{24!}{21!\cdot3!}=\binom{24}{3}\), and the second term is \(\frac{23!}{20!\cdot3!}=\binom{23}{3}\). By Pascal’s identity, \(\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}\), the expression becomes \(\binom{23}{2}\). Thus, \(\binom{23}{2}=\frac{23\times22}{2}=253\), so option C is correct. The value 252 may result from an incorrect calculation after applying the identity. Exam tip: in differences of consecutive binomial coefficients, look for Pascal’s identity first.

The expression is \(\frac{15!}{11!\cdot4!}\times\frac{10!\cdot4!}{14!}\). After cancelling \(4!\), it becomes \(\frac{15!\cdot10!}{11!\cdot14!}\). Using \(15!=15\cdot14!\) and \(11!=11\cdot10!\), all common factors cancel and the value is \(\frac{15}{11}\). \(\frac{14}{11}\) could result from incorrectly omitting the extra factor 15 in \(15!\). Exam tip: Expand factorials only as far as needed to cancel common factorial terms.

Expanding the numerator down to the denominator factorial gives \((2n+3)!=(2n+3)(2n+2)(2n+1)(2n)(2n-1)!\). Cancelling \((2n-1)!\) leaves four factors: \((2n+3),(2n+2),(2n+1)\), and \(2n\). Hence, the correct answer is 4. Option 5 is incorrect because the difference between the factorial indices is \((2n+3)-(2n-1)=4\). Exam tip: in \(a!/b!\), where \(a>b\), the number of factors left is \(a-b\).

Cancelling the common factorial terms gives \(\frac{(2n+3)!}{(2n-1)!}=(2n)(2n+1)(2n+2)(2n+3)\). Since \(17160=10\times11\times12\times13\), we have \(2n=10\), and hence \(n=5\). For option 4, the product is \(8\times9\times10\times11\), which is not 17160. Exam tip: In a factorial ratio, expand only the consecutive factors that remain after the smaller factorial is cancelled.

Expanding the numerator up to the factorial in the denominator gives \((4n+1)!=(4n+1)(4n)(4n-1)!\). The factor \((4n-1)!\) then cancels, leaving \((4n+1)(4n)\). Option D incorrectly retains the factor \((4n-1)\). Exam tip: In a factorial quotient, expand the larger factorial only as far as the denominator factorial.

Write \((4n+1)!=(4n+1)(4n)(4n-1)!\). Cancelling \((4n-1)!\) gives \((4n+1)(4n)=420\). Since \(20\times21=420\), we get \(4n=20\), hence \(n=5\). The nearest distractor, \(n=4\), gives \(16\times17=272\), not 420. Exam tip: for a quotient of nearby factorials, expand the larger factorial only up to the smaller one and cancel common factors.

The simplified form is (3(n+5)(n+4)). Since (3\cdot13\cdot12=468) is wrong and (3\cdot14\cdot13=546), (n=8).

\(\frac{(n+6)!}{(n+4)!}=(n+6)(n+5)\) and \(\frac{(n+5)!}{(n+3)!}=(n+5)(n+4)\). Hence, the sum is \((n+5)[(n+6)+(n+4)]\). This gives \((n+5)(2n+10)=2(n+5)^2\), so option A is correct. Option B has \(2n+9\), whereas the correct sum inside the bracket is \(2n+10\). Exam tip: In factorial division, cancel through the denominator factorial and multiply the remaining consecutive factors.

\(\frac{(n+6)!}{(n+4)!}=(n+6)(n+5)\) and \(\frac{(n+5)!}{(n+3)!}=(n+5)(n+4)\). Thus, their sum is \((n+5)[(n+6)+(n+4)]=2(n+5)^2\). Hence \(2(n+5)^2=338\), so \((n+5)^2=169\). Since the factorial arguments must be valid integers, take \(n+5=13\), which gives \(n=8\). Substituting \(n=9\) gives \(392\), not \(338\). Exam tip: In factorial fractions, cancel the common factorial part first and write the remaining consecutive factors.

The first term is \(\frac{13!}{7!\cdot6!}=\binom{13}{6}=1716\), and the second term is \(\frac{13!}{8!\cdot5!}=\binom{13}{5}=1287\). Hence, the required value is \(1716+1287=3003\). The value \(3432\) is twice \(1716\), so it is not the required sum. Exam tip: Recognising \(\frac{n!}{r!(n-r)!}\) as \(\binom{n}{r}\) makes factorial questions faster to solve.

\(\frac{12!}{6!\cdot6!}=\binom{12}{6}=924\). Therefore, \(924\times\frac{3}{11}=84\times3=252\). Hence, 252 is correct. Getting 264 would require incorrectly taking \(924\div11\) as 88, whereas it equals 84. Exam tip: first recognise such factorial ratios as \(\binom{n}{r}\), then simplify.

Factorial quotients are easiest to evaluate by expanding only the factors that do not cancel. Since \(14!=14\times13\times12\times11!\), dividing by \(11!\) leaves \(14\times13\times12\). The same idea applies to the second quotient, allowing the expression to be evaluated without large factorials.

The first term is \(\frac{14!}{11!}=14\times13\times12=2184\). The second part is \(3\cdot\frac{13!}{11!}=3\cdot(13\times12)=3\cdot156=468\). Subtracting gives \(2184-468=1716\). Therefore option B is correct. The factor 3 multiplies the entire second quotient; it is not part of the factorial denominator and must be applied after simplifying that quotient.

Using \(10!=10\cdot9\cdot8!\) and \(9!=9\cdot8!\), we get \(\frac{10!+3\cdot9!}{8!}=\frac{10\cdot9\cdot8!+3\cdot9\cdot8!}{8!}=90+27=117\). Hence, 117 is correct. A result such as 126 may arise from multiplying the \(3\cdot9!\) term incorrectly. Exam tip: express every factorial in terms of the factorial in the denominator, here \(8!\), before simplifying.

Expand \((n+4)!\) down to \((n-1)!\): \((n+4)!=(n+4)(n+3)(n+2)(n+1)n(n-1)!\). In the given expression, \((n-1)!\), \((n+3)\), and \((n+2)\) cancel, leaving \(n(n+1)(n+4)\). Hence, option A is correct. Option B incorrectly keeps \((n+2)\), although it cancels with the denominator. Exam tip: in factorial fractions, expand the larger factorial only up to the smaller factorial before cancelling.

Using \((n+4)!=(n+4)(n+3)(n+2)(n+1)n(n-1)!\), the expression simplifies to \(n(n+1)(n+4)\). At \(n=7\), it equals \(7\times8\times11=616\), and at \(n=8\), it equals \(8\times9\times12=864\). Since \(n(n+1)(n+4)\) increases for natural numbers, it cannot take the value 840 between these two consecutive cases. Hence, there is no natural-number solution. Exam tip: Expand factorials into consecutive factors first, then cancel common terms.

The exponent is ( \left\lfloor\frac{35}{3}\right\rfloor+\left\lfloor\frac{35}{9}\right\rfloor+\left\lfloor\frac{35}{27}\right\rfloor=15 ). Do not forget the contribution of higher powers.

(12!) has exponent (10) of (2) and exponent (5) of (3). Therefore all given conditions are first satisfied by (12!).