Factorial notation

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

Expanding \((n+1)!\) up to \((n-2)!\) gives \((n+1)!=(n+1)n(n-1)(n-2)!\). Cancelling \((n-2)!\), we get \(\frac{(n+1)!}{(n-2)!}=(n+1)n(n-1)\). Option B incorrectly retains an extra factor \((n-2)\). Exam tip: expand the numerator factorial only until the denominator factorial appears, then cancel common factors.

Correct answer: B. (330)

Each term has the form \(\frac{n!}{r!(n-r)!}\), which is the combination \(\binom{n}{r}\). Thus the first term is \(\binom{10}{4}=210\), and the second is \(\binom{10}{3}=120\). Recognising this form makes the calculation shorter, although the same result follows by cancelling factorials directly.

Adding the two separately evaluated terms gives \(210+120=330\). Therefore option B is correct. It is important not to add the denominators or combine the numerators before evaluating, because factorial expressions do not distribute over addition in that way. The values 240, 360, and 450 do not result from the correct combination calculations.

Correct answer: A. 18

Since \(18! = 18\times17\times16!\), we get \(\frac{18!}{16!\cdot2!}=\frac{18\times17}{2}=153\). Therefore, \(153\times\frac{2}{17}=18\). Hence, the correct answer is 18. Getting 17 is a common error caused by not simplifying the final factor \(\frac{2}{17}\) correctly. Exam tip: In factorial questions, expand the larger factorial only up to the smaller factorial and cancel common factors first.

Correct answer: D. (13)

After canceling common (10!), the ratio becomes ( \frac{13!}{12!}=13 ). You can also check by inverting the second fraction.

Correct answer: C. (8)

((n+4)(n+3)(n+2)=990), and (11\cdot10\cdot9=990), so (n=7). Match the three consecutive factors carefully.

Correct answer: A. (132)

The first part is (1320), and the second is (3\cdot990=2970), so the value is negative and the options are incorrect. Always check signs and multipliers.

Correct answer: A. (91)

( \frac{15!}{12!}=15\cdot14\cdot13 ) and ( \frac{5!}{2!}=60 ), so the value is (45.5); the options are incorrect. Do not force-fit an answer.

Correct answer: B. (49)

Both fractions can be recognised as combinations. The first is \(\frac{7!}{2!5!}=\binom{7}{2}=\frac{7\times6}{2}=21\). The second is \(\frac{8!}{2!6!}=\binom{8}{2}=\frac{8 imes7}{2}=28\). Evaluating each term separately prevents confusion between the two different numerators and denominators.

Their sum is \(21+28=49\), so option B is correct. Another useful observation is that these are consecutive values of choosing two objects: \(\binom{7}{2}\) and \(\binom{8}{2}\). They must be added as complete values; adding only selected factors or factorial denominators would not represent the original expression. The supplied answer and calculation are therefore consistent.

Correct answer: D. (3(n+1)(n+2))

The first term is ((n+3)(n+2)(n+1)) and the second is (n(n+1)(n+2)), so the difference is (3(n+1)(n+2)). Take common factors out.

Correct answer: A. (6)

The simplified form is (3(n+1)(n+2)), so ((n+1)(n+2)=72) and (n=7); the correct option is (7). First simplify and then match options.

Correct answer: C. (120)

(11!=11\cdot10\cdot9!) and (10!=10\cdot9!), so the value is (120). Taking the common factorial in addition is a quick method.

Correct answer: B. 8

\(\frac{(n+2)!}{n!}=(n+2)(n+1)\) and \(\frac{(n+1)!}{(n-1)!}=n(n+1)\). Hence, their difference is \((n+1)[(n+2)-n]=2(n+1)\). So \(2(n+1)=18\), giving \(n+1=9\) and therefore \(n=8\). Option 9 results from incorrectly taking \(n+1\) itself as \(n\). Exam tip: In factorial fractions, cancel common factorial terms first and then expand the remaining factors.

Correct answer: A. (28)

The first part is (364), and (364\cdot\frac{3}{13}=84). Write each cancellation clearly during calculation.

Correct answer: B. ( \frac{12}{5} )

( \frac{9!}{7!}=72 ) and ( \frac{4!}{6!}=\frac{1}{30} ), so the value is ( \frac{12}{5} ). Break large products into smaller ratios.

Correct answer: B. ( (2n+1)(2n) )

((2n+1)!=(2n+1)(2n)(2n-1)!), so the answer is ((2n+1)(2n)). Do not skip the order in indexed factorials.

Correct answer: B. (84)

The first term is (210) and the second is (84), so the difference is (126); the correct option should be (126). Check the final subtraction before choosing an option.

Correct answer: C. (1320)

Writing (12!) as (12\cdot11\cdot10\cdot9!) gives (1320). In exams, expand the larger factorial up to the smaller factorial.

Correct answer: B. (120)

Writing (10!) as (10\cdot9\cdot8\cdot7!) and dividing by (3!) gives (120). Cancel common factorials first.

Correct answer: A. (8)

It gives ((n+2)(n+1)=90), and (10\cdot9=90), so (n=8). Recognizing consecutive factors is a quick method.

Correct answer: C. (8)

The direct answer is C, \(n=8\). Expand the factorial quotient: \(\frac{n!}{(n-3)!}=n(n-1)(n-2)\). Hence we need three consecutive decreasing positive integers whose product is 336. Test the useful candidate from the options: \(8\times7\times6=56\times6=336\), so \(n=8\). Option A, 6, gives \(6\times5\times4=120\), not 336. Option B, 7, gives \(7\times6\times5=210\), not 336. Option C, 8, gives exactly 336 and is correct. Option D, 9, gives \(9\times8\times7=504\), too large. The factorial quotient requires three factors, not just one number. Memory cue: cancel \((n-3)!\) from \(n!\), leaving the three descending terms \(n,n-1,n-2\).

Correct answer: B. ( (n+1)(n+3) )

((n+1)!) is common in both terms, so the simplified form is ((n+1)(n+3)). Take the common factorial out first.

Correct answer: D. 5

The governing concept is factorial notation and cancellation. Use (n+3)! = (n+3)(n+2)! to factor the numerator: (n+3)! − (n+2)! = (n+2)![(n+3) − 1] = (n+2)!(n+2). Dividing by (n+1)! gives (n+2)!/(n+1)! × (n+2) = (n+2)(n+2) = (n+2)^2. Hence (n+2)^2 = 49. In the usual school definition of factorial, the relevant factorial arguments must be non-negative integers, so n+2 is positive. Therefore n+2 = 7 and n = 5, making option D correct. The algebraic root n = −9 is not admissible in this factorial setting. Options A, B, and C produce values different from 49.

Correct answer: A. ( \frac{52}{5} )

The value is ( \frac{14\cdot13\cdot12}{7\cdot6\cdot5}=\frac{52}{5} ). In division, expand both factorial ratios separately.

Correct answer: C. 252

Since \(9!=9\times8\times7\times6!\), \(\frac{9!}{6!\cdot2!}=\frac{9\times8\times7}{2}=252\). Therefore, 252 is correct. The value 336 is a common error obtained by not dividing by \(2!\). Exam tip: In factorial fractions, cancel the common factorial term first and then multiply the remaining factors.

Correct answer: B. ( (2n+2)(2n+1) )

((2n+2)!=(2n+2)(2n+1)(2n)!), so the answer is ((2n+2)(2n+1)). Do not skip the order in indexed factorials.