Factorial notation

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.

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.

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.

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

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

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.

( \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.

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.

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.

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.

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

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

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

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

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

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.

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

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

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

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 \(n,n-1,n-2\).

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

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.

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

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.

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