Factorial notation

( \frac{4!}{5!}=\frac{1}{5} ) and ( \frac{8!}{7!}=8 ), so the value is ( \frac{8}{5} ). Cancel separate ratios in multiplication.

\(\frac{(n+1)!}{(n-1)!}=n(n+1)\), while \(\frac{n!}{(n-2)!}=n(n-1)\). Therefore, the difference is \(n(n+1)-n(n-1)=n[(n+1)-(n-1)]=2n\). \(2n+1\) would result from an incorrect simplification of the subtraction. Exam tip: expand factorials only up to the denominator factorial and cancel common factors.

\(\frac{(n+1)!}{(n-1)!}=n(n+1)\) and \(\frac{n!}{(n-2)!}=n(n-1)\). Their difference is therefore \(n(n+1)-n(n-1)=2n\). Hence \(2n=18\), giving \(n=9\). For example, if \(n=10\), the difference would be \(20\), not 18. Exam tip: expand factorials only up to the common smaller factorial, then cancel.

\(\frac{(n+2)!}{(n-1)!}=n(n+1)(n+2)\), while \(\frac{(n+1)!}{(n-2)!}=(n-1)n(n+1)\). Therefore, the difference is \(n(n+1)[(n+2)-(n-1)] = 3n(n+1)\). Option C is only the first term, so it is not correct. Exam tip: expand the numerator factorial down to the denominator factorial, then factor out common terms.

\(\frac{(n+2)!}{(n-1)!}=n(n+1)(n+2)\) and \(\frac{(n+1)!}{(n-2)!}=(n-1)n(n+1)\). Their difference is \(n(n+1)[(n+2)-(n-1)]=3n(n+1)\). Hence, \(3n(n+1)=216\), so \(n(n+1)=72\). Since \(8\times9=72\), \(n=8\). For \(n=7\), the product is only \(56\). Exam tip: In factorial fractions, cancel common terms and rewrite the result as consecutive factors.

The expansion is ((n+5)(n+4)(n+3)(n+2)), so (4) factors remain. The factorial gap gives the number of factors.

The direct answer is option C, 1680. Substitute \(n=3\): \(n+5=8\) and \(n+1=4\). Thus \(\frac{(n+5)!}{(n+1)!}=\frac{8!}{4!}\). Cancel \(4!\) from the factorials: \(\frac{8!}{4!}=8\times7\times6\times5=1680\). Option C is correct. Option A, 840, is exactly half of the correct product and may result from forgetting one factor or mishandling \(2!\)-like cancellation. Option B, 1260, does not equal the required consecutive product. Option D, 2016, is also not the value of \(8\times7\times6\times5\). The safe rule is \(\frac{m!}{r!}=m(m-1)\cdots(r+1)\); after substituting, write every remaining factor before multiplying.

Since \(20!=20\cdot19\cdot18!\) and \(2!=2\), the first term is \(\frac{20!}{18!\cdot2!}=\frac{20\cdot19}{2}=190\). Also, \(\frac{19!}{18!}=19\). Hence, the value of the expression is \(190-19=171\). Option 190 represents only the first term, not the complete expression. Exam tip: In factorial fractions, first cancel the common factorial terms before calculating.

The two terms are (462) and (252), so the ratio is ( \frac{11}{6} ). Reduce large values to a simple ratio.

Cancelling \((n+2)!\) from \((n+4)!\) gives \(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\). Hence, \((n+4)(n+3)=182\). Since \(182=14\times13\), we get \(n+4=14\), so \(n=10\). For option 9, the product would be \(13\times12=156\), so it is not correct. Exam tip: In a factorial fraction, expand the larger factorial only as far as needed and cancel the common factorial term.

Writing (16!) as (16\cdot15\cdot14\cdot13!) gives (3360). In exams, expand the larger factorial up to the smaller factorial.

( \frac{13\cdot12\cdot11}{3!}=\frac{1716}{6}=286 ). Replace the small factorial in the denominator correctly.

(12!-11!=11!(12-1)) and ( \frac{11!}{10!}=11 ), so the value is (121). Always take a common factor in subtraction.

It gives \((n+2)(n+1)=132\), and \(12\cdot11=132\), so \(n=10\). Recognize consecutive factors.

Direct answer: Option C, \\(n=9\\). Expand the factorial ratio by cancelling \\( (n-3)!\\): \\(n!/(n-3)!=n(n-1)(n-2)\\). Therefore solve \\(n(n-1)(n-2)=504\\). Testing consecutive factors, \\(9\cdot8\cdot7=72\cdot7=504\\), so n=9. Option A, n=7, gives \\(7\cdot6\cdot5=210\\), not 504. Option B, n=8, gives \\(8\cdot7\cdot6=336\\), not 504. Option C gives exactly 504 and is positive. Option D, n=10, gives \\(10\cdot9\cdot8=720\\), too large. Factorial ratios are products of consecutive descending integers, so looking near the cube root of 504 is useful; \\(8^3=512\\), which points to 9, 8, 7. Memory cue: cancel factorials before calculating.

Direct answer: Option B, \\(n(n+2)\\). Rewrite both factorials using n!: \\( (n+1)!=(n+1)n!\\), and \\(n!=n(n-1)!\\). The numerator becomes \\( (n+1)!+n!=(n+1)n!+n!=n![(n+1)+1]=n!(n+2)\\). Dividing by \\( (n-1)!\\), and using \\(n!/(n-1)!=n\\), gives \\(n(n+2)\\). Option A, n^2, misses the extra 2n term. Option B matches the derived result. Option C, n(n+1), would come from incorrectly treating the numerator as only (n+1)!, ignoring n!. Option D, n+2, omits the factor n from the factorial ratio. The safest method is to take n! common before cancelling. This also requires the usual factorial domain where n is a positive integer.

Using \((n+1)!=(n+1)n!\) and \(n!=n(n-1)!\),
\[
\frac{(n+1)!+n!}{(n-1)!}=\frac{n![(n+1)+1]}{(n-1)!}=n(n+2).
\]
Thus, \(n(n+2)=63\), or \(n^2+2n-63=0\). This gives \((n-7)(n+9)=0\). Since a factorial is not defined for a negative integer, \(n=-9\) is invalid; hence \(n=7\). Exam tip: in factorial fractions, first express terms using the smallest factorial present, here \((n-1)!\), and then cancel.

Using \((n+2)!=(n+2)(n+1)!\), the numerator becomes \((n+1)![(n+2)-1]=(n+1)!(n+1)\). On dividing by \(n!\), we get \((n+1)^2=100\). Since \(n\) must be a non-negative integer for factorial notation, \(n+1=10\), so \(n=9\). For option 10, the value would be \(121\), not 100. Exam tip: In factorial expressions, factor out the smaller factorial first for quick simplification.

( \frac{9!}{7!}=72 ) and ( \frac{5!}{6!}=\frac{1}{6} ), so the value is (12). Cancel separate ratios in multiplication.

The first ratio is (210) and the second is (120), so the quotient is ( \frac{7}{4} ). Solve division as a ratio.

First simplify the factorial quotient by cancellation: 15!/(12!·3!) = (13 × 14 × 15)/6 = 2730/6 = 455. Now multiply by 2/13: 455 × 2/13 = (35 × 13) × 2/13 = 70. Therefore option D is correct. The original option list did not contain the mathematically correct value 70, so option D has been corrected from 40 to 70 to preserve one unambiguous answer. Options A, B, and C are not obtained from the stated expression; they may result from incomplete cancellation or an arithmetic mistake. The factorial identity and cancellation are the governing concepts.

The expression uses factorial cancellation and then a simple multiplication. A factorial such as \(15!\) contains all factors from 1 through 15, while \(12!\) contains those from 1 through 12. Dividing them leaves only \(13\times14\times15\). The factor \(3!\) must also be included in the denominator, so the first part is \(\frac{13\times14\times15}{6}=455\). This makes the calculation shorter and less error-prone.

Now multiply by \(\frac{3}{13}\). Cancelling 13 first gives \(\frac{14\times15\times3}{6}=14\times15\times\frac12=105\). Therefore the value is 105, which is option A. The other numerical options result from incomplete cancellation or incorrect factorial arithmetic.

The first term is (153) and the second is (136), so the difference is (17). Subtract nearby factorial terms carefully.

It is ((2n+1)(2n)=306), and no integer (n) satisfies it. Check whether the matched consecutive factors fit the form.

It is ((2n+1)(2n)=272), and (17\cdot16=272), so (2n=16) and (n=8). Identify the consecutive factors first.