Factorial notation

The \(n!\) factors cancel in the given complex fraction:
\(\frac{\frac{(n+4)!}{n!}}{\frac{(n+2)!}{n!}}=\frac{(n+4)!}{(n+2)!}\).
Since \((n+4)!=(n+4)(n+3)(n+2)!\), the factor \((n+2)!\) cancels, leaving \((n+4)(n+3)\). Option C does not result from cancelling the factorials correctly. Exam tip: expand the larger factorial only up to the smaller factorial before cancelling common factors.

The recursive factorial property is \(n!=n(n-1)!\) for a positive integer \(n\). Also, \(0!=1\), so option B is false, and factorial is not defined for negative integers. Exam tip: remember factorial as a product, not a sum.

On cancelling \((n-2)!\) from \((n+1)!\), we get \(\frac{(n+1)!}{(n-2)!}=(n+1)n(n-1)\). Hence, \((n+1)n(n-1)=60\). For \(n=4\), \(5\times4\times3=60\), so the correct value is \(\boxed{4}\). The nearby option \(n=3\) gives \(4\times3\times2=24\), not 60. Exam tip: In factorial fractions, expand the numerator factorial up to the denominator factorial and cancel common factors.

Since \(n!\) means \(n\times(n-1)\times\cdots\times1\), it follows that \(n!=n(n-1)!\). Option B incorrectly divides by \(n\), reversing the relation. Exam tip: when reducing a factorial by one, check that the factor \(n\) is multiplied.

Since \((n+3)!=(n+3)(n+2)!\), the numerator becomes \((n+2)![(n+3)+1]\). Dividing by \((n+2)!\) gives \(n+4\). Choosing \(n+3\) would ignore the contribution of the second term, \((n+2)!\), in the numerator. Exam tip: In factorial expressions, first rewrite the larger factorial as a multiple of the smaller one.

Since \(10!=10\times9\times8!\), \(\frac{10!}{8!}=10\times9=90\). Therefore, \(\frac{10!-8!}{8!}=\frac{10!}{8!}-\frac{8!}{8!}=90-1=89\). The close option \(90\) results from forgetting to subtract the \(8!\) term. Exam tip: In factorial fractions, cancel common factorial terms first.

\(4!=24\) and \(3!=6\). Also, \(\frac{6!}{5!}=\frac{6\times5!}{5!}=6\). Therefore, the expression becomes \(24\times6-6=144-6=138\). Option D, 144, results if \(3!\) is not subtracted at the end. Exam tip: For a quotient of consecutive factorials, write the larger factorial in terms of the smaller one and cancel common factors.

\(\frac{11!}{8!\,3!}=\frac{11\times10\times9}{3\times2\times1}=165\), and \(\frac{9!}{7!\,2!}=\frac{9\times8}{2\times1}=36\). Therefore, the required value is \(165-36=129\). A value such as \(126\) can result from simplifying the first term incorrectly. Exam tip: cancel common factorial parts first, then multiply the remaining factors.

Expanding the numerator gives \((n+5)!=(n+5)(n+4)(n+3)(n+2)!\). Cancelling \((n+2)!\) therefore leaves \((n+5)(n+4)(n+3)\). Option A misses the factor \((n+3)\). Exam tip: Expand the larger factorial only up to the factorial in the denominator, then cancel the common factor.

Cancelling the common factorial terms gives \(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\). Hence, \((n+4)(n+3)=132\). The consecutive factors of 132 are \(11\) and \(12\), so \(n+3=11\) and therefore \(n=8\). If \(n=7\), the product is \(11\times10=110\), so it is not correct. Exam tip: In a factorial fraction, expand only up to the smaller factorial and cancel common terms.

\(\frac{7!}{5!}=7\times6=42\), since \(5!\) cancels out. Also, \(\frac{5!}{2!\,3!}=\frac{120}{2\times6}=10\), and \(1!=1\). Therefore, the total is \(42+10+1=53\). One may get \(52\) by incorrectly taking \(1!\) as 0, but \(1!=1\). Exam tip: simplify factorial fractions by cancelling common factorial terms before calculating.

Since \((n+2)!=(n+2)(n+1)!\), \(\frac{(n+2)!}{(n+1)!}=n+2\). Similarly, \((n+1)!=(n+1)n!\), so \(\frac{(n+1)!}{n!}=n+1\). Therefore, the complete expression is \((n+2)(n+1)\). The option \(n+2\) represents only the first ratio and misses the factor \(n+1\). Exam tip: In factorial ratios, expand the larger factorial up to the smaller one and cancel common factors.

The factorial terms in the given expression are multiplied: \(\frac{8!\times3!}{6!\times2!}\). Now, \(\frac{8!}{6!}=8\times7=56\) and \(\frac{3!}{2!}=3\). Hence, the value is \(56\times3=168\). A value such as 196 results from not simplifying the factorials correctly. Exam tip: Cancel common factorial parts first and then multiply the remaining factors.

Since \(9!=9\times8\times7!\), \(\frac{9!}{7!}=9\times8=72\). Therefore, \(\frac{9!-7!}{7!}=\frac{9!}{7!}-\frac{7!}{7!}=72-1=71\). Option 72 is only the value of \(\frac{9!}{7!}\); subtracting \(\frac{7!}{7!}=1\) is still required. Exam tip: In factorial fractions, expand the larger factorial up to the smaller factorial before cancelling common factors.

Using \(n!=n(n-1)!\), the numerator becomes \(n(n-1)!+(n-1)!=(n-1)!(n+1)\). Dividing by \((n-1)!\) gives \(n+1\). Option \(n\) would result from incorrectly ignoring the second term \((n-1)!\). Exam tip: first factor out the common \((n-1)!\) in factorial expressions.

Factorials can be simplified by cancelling the common consecutive factors. Substituting n = 5 gives \\(\\frac{8!}{6!}-\\frac{6!}{5!}\\). Since \\(8!=8\\times7\\times6!\\), the first fraction equals \\(8\\times7=56\\). Also, \\(6!=6\\times5!\\), so the second fraction equals 6. The subtraction must be done after simplifying both fractions.

Therefore the required value is \\(56-6=50\\). This corresponds to option C, not option D. The result can also be checked directly from the general expression: \\(\\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\\) and \\(\\frac{(n+1)!}{n!}=n+1\\); for n = 5 this gives 8 × 7 − 6 = 50. Thus the supplied answer C is correct.

\(\frac{6!}{2!\,4!}=\frac{6\times5}{2\times1}=15\), and \(\frac{4!}{3!}=4\). Therefore, the given expression is \(15\times4=60\). Do not cancel \(4!\) casually just because it appears in both fractions; first combine the product into a single numerator and denominator. Exam tip: While simplifying factorials, write the full product clearly before cancelling common factors.

The governing concept is cancellation in factorial ratios. Since 14! = 14 × 13 × 12!, the first ratio is 14!/12! = 14 × 13 = 182. Likewise, 13! = 13 × 12 × 11!, so the second ratio is 13!/11! = 13 × 12 = 156. Subtracting gives 182 - 156 = 26. Therefore option B is correct. A useful check is to factor the difference: 14 × 13 - 13 × 12 = 13(14 - 12) = 13 × 2 = 26. Option A could arise from an incomplete subtraction, while options C and D do not agree with either the direct calculation or the factored form. The factorials should not be evaluated as enormous full products; only the uncancelled factors are needed.

Write \((n+2)!=(n+2)(n+1)!\). The numerator becomes \((n+1)!\big[(n+2)-1\big]=(n+1)!(n+1)\). Since \((n+1)!=(n+1)n!\), the given expression simplifies to \((n+1)^2\). Thus, \((n+1)^2=36\), giving \(n+1=6\) and \(n=5\). The other algebraic value, \(n=-7\), is not valid because factorials of negative integers are undefined. Exam tip: Express factorial terms in terms of the smallest factorial before cancelling.

The \(n!\) in the denominators of the two inner fractions cancels, giving \(\frac{(n+3)!}{(n+2)!}\). Since \((n+3)!=(n+3)(n+2)!\), this becomes \(n+3\). Therefore, the correct answer is \(n+3\). The expression \(\frac{n+3}{n+2}\) results from the incorrect assumption that \((n+2)!\) equals \(n+2\). Exam tip: In factorial division, expand the larger factorial only up to the smaller factorial and then cancel.

\(\frac{10!}{6!\,4!}=\binom{10}{4}=210\) and \(\frac{10!}{7!\,3!}=\binom{10}{3}=120\). Therefore, their sum is \(210+120=330\), so option D is correct. Exam tip: Recognising expressions of the form \(\frac{n!}{r!(n-r)!}\) as combinations makes calculation quicker.

The direct answer is option B, 2729. Use factorial cancellation instead of calculating enormous factorials. Divide each term in the numerator by \(12!\): \(\frac{15!-12!}{12!}=\frac{15!}{12!}-\frac{12!}{12!}\). Now \(\frac{15!}{12!}=15\times14\times13=2730\), while \(\frac{12!}{12!}=1\). Therefore the result is \(2730-1=2729\). Option B is correct. Option A, 2728, is one less than the correct value and can result from an arithmetic error. Option C, 2730, is only the value of \(\frac{15!}{12!}\); it forgets the subtraction of 1. Option D, 2731, is too large and does not follow from the subtraction. The useful rule is \(\frac{m!}{r!}=m(m-1)\cdots(r+1)\) when \(m>r\), and a factorial divided by itself equals 1.

Cancelling \((n+1)!\) after expanding \((n+4)!\) gives \(\frac{(n+4)!}{(n+1)!}=(n+4)(n+3)(n+2)\). Hence, \((n+4)(n+3)(n+2)=504\). Since \(7\times8\times9=504\), we get \(n+2=7\), so \(n=5\). For example, \(n=4\) gives \(6\times7\times8=336\), not \(504\). Exam tip: In factorial fractions, expand the larger factorial only up to the smaller factorial and cancel common terms.

\(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\) and \(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\). Therefore, the expression becomes \((n+3)[(n+4)-(n+2)] = 2(n+3)\). \(2n+4\) is incorrect because \(2(n+3)=2n+6\). Exam tip: In factorial quotients, cancel the common factorial part and write the remaining consecutive factors.

By the definition of factorial, \(8!=8\cdot7\cdot6\cdot5\cdots1\), while \(6!=6\cdot5\cdots1\). Hence, \(8!=8\cdot7\cdot6!\), so option A is correct. In \(8\cdot6!\), the factor \(7\) is missing. Exam tip: When expressing \(n!\) in terms of \((n-r)!\), include every intervening factor.