Factorial notation

Correct answer: D. ( \frac{8}{5} )

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

Correct answer: C. \(2n\)

\(\frac{(n+1)!}{(n-1)!}=n(n+1)\), 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.

Correct answer: B. 9

\(\frac{(n+1)!}{(n-1)!}=n(n+1)\) 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.

Correct answer: A. \(3n(n+1)\)

\(\frac{(n+2)!}{(n-1)!}=n(n+1)(n+2)\), 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.

Correct answer: D. 8

\(\frac{(n+2)!}{(n-1)!}=n(n+1)(n+2)\) 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.

Correct answer: B. (4)

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

Correct answer: C. (1680)

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.

Correct answer: A. 171

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.

Correct answer: D. ( \frac{11}{6} )

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

Correct answer: B. 10

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.

Correct answer: C. (3360)

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

Correct answer: B. (286)

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

Correct answer: A. (121)

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

Correct answer: C. (10)

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

Correct answer: C. (9)

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.

Correct answer: B. (n(n+2))

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.

Correct answer: C. 7

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.

Correct answer: C. 9

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.

Correct answer: A. (12)

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

Correct answer: B. ( \frac{7}{4} )

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

Correct answer: D. 70

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.

Correct answer: A. (105)

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.

Correct answer: C. (17)

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

Correct answer: B. (8)

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

Correct answer: C. (8)

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