Factorial notation

A quotient of factorials can usually be simplified by expanding only the numerator until the denominator appears. This reveals a common factorial factor that cancels. Such cancellation is safer and faster than computing 8! and 6! separately, especially because factorial values grow rapidly.

Since \(8!=8\times7\times6!\), we have \(\frac{8!}{6!}=\frac{8\times7\times6!}{6!}=8\times7=56\). Therefore option A is correct. The value 48 would come from an incorrect product, 14 from adding rather than multiplying, and 336 from failing to cancel or using an unrelated calculation. The expression is a factorial quotient, not a sum or ordinary difference.

Since \((n+1)!=(n+1)n(n-1)!\), we get \(\frac{(n+1)!}{(n-1)!}=n(n+1)\). Hence, \(n(n+1)=42\), or \(n^2+n-42=0\). The positive solution is \(n=6\), because \(6\times7=42\). Option 7 is not correct because it gives \(7\times8=56\). Exam tip: In factorial ratios, first cancel the common factorial part and write the remaining consecutive factors.

Since \(7!=7\times6!\), \(\frac{7!+6!}{6!}=\frac{7\times6!+6!}{6!}=\frac{8\times6!}{6!}=8\). Option 7 is only the value of \(7!/6!\); it ignores the additional \(6!\) term. Exam tip: in a sum of factorial terms, first factor out the smaller common factorial.

The direct answer is A, 22. Expand only the part needed for cancellation: \(12!=12\times11\times10!\), and \(3!=3\times2\times1=6\). Thus \(\frac{12!}{10!\cdot3!}=\frac{12\times11\times10!}{10!\times6}=\frac{12\times11}{6}=\frac{132}{6}=22\). Option A, 22, is correct because the common \(10!\) cancels and the remaining numerator is divided by 6. Option B, 24, is wrong because it does not correctly divide 132 by 6. Option C, 132, is the value before dividing by \(3!=6\), so it is incomplete. Option D, 220, is wrong and does not follow from the factorial simplification. A common mistake is to cancel \(10!\) but forget the additional denominator factor \(3!\).

Cancelling \((n-2)!\) gives \(\frac{n!}{(n-2)!}=n(n-1)\). Hence, \(n(n-1)=90\). Since \(90=10\times9\), the required consecutive positive integers are 10 and 9, so \(n=10\). For option 9, the product would be \(9\times8=72\), not 90. Exam tip: In such questions, expand the factorial first and cancel the common factorial terms.

Writing (9!) as (9\cdot8\cdot7\cdot6\cdot5!) and dividing by (4!) gives (126). This is a common form in combinations.

((n+3)!=(n+3)(n+2)(n+1)!), so the answer is ((n+3)(n+2)). Expand the larger factorial up to the smaller one.

Using (6!) and (5!=5\cdot4!), the value becomes ( \frac{5}{7} ). In products, cancel smaller factorials carefully.

The first term is (56) and the second term is (35), so the sum is (91). Simplify each term separately in mixed expressions.

Since \((n+4)!=(n+4)(n+3)(n+2)!\), dividing by \((n+2)!\) cancels the common factorial and gives \(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\). Option A incorrectly retains an extra factor \((n+2)\). Exam tip: Expand the larger factorial only up to the smaller factorial, then cancel the common part.

On cancelling \((n+1)!\) from \((n+3)!\), we get \(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\). Hence, \((n+3)(n+2)=72\). The consecutive factors of 72 are \(8\) and \(9\), so \(n+2=8\) and \(n=6\). For option 7, the product would be \(9\times10=90\), not 72. Exam tip: In a factorial fraction, expand the larger factorial only up to the smaller one and cancel common factors.

The first part is (84) and the second part is (3), so the product is (252). Observe the structure carefully before canceling.

The direct answer is B, \(2n(2n-1)\). Expand the numerator only until the denominator appears: \((2n)!=(2n)(2n-1)(2n-2)!\). Therefore, \(\frac{(2n)!}{(2n-2)!}=\frac{(2n)(2n-1)(2n-2)!}{(2n-2)!}=2n(2n-1)\). Option A, \(2n\), omits one remaining factor, namely \(2n-1\). Option B is correct because exactly two factors remain after cancellation. Option C, \(2n(2n+1)\), uses a factor larger than the first numerator factor and does not match factorial order. Option D, \((2n-1)(2n-2)\), starts one step too low and includes a factor that cancels. The exam rule is: when factorials differ by two, the answer is the product of the two descending factors beginning with the larger top number.

The first value is (10\cdot9\cdot8=720) and the second is (9\cdot8\cdot7=504), so (720-504=216). Check subtraction carefully after expansion.

(5!=20\cdot3!), (4!=4\cdot3!), and (3!=1\cdot3!), so the total is (25). Write every term in the same factorial form.

Use the factorial identity \\(n!=n(n-1)!\\). The numerator is \\((n+1)!-n!= (n+1)n!-n! = n\,n!\\). Since \\(n!=n(n-1)!\\), this becomes \\(n^2(n-1)!\\). Dividing by \\((n-1)!\\), the expression simplifies to \\(n^2\\), provided the factorials are defined. Thus \\(n^2=49\\), giving \\(n=7\\) or \\(n=-7\\) algebraically.

In factorial notation for this school problem, \\(n\\) is a nonnegative integer, and the denominator requires \\(n\\ge1\\). Therefore only \\(n=7\\) is allowed. Option C is correct. The negative root is rejected because negative integers do not have the required ordinary factorial interpretation here.

The direct answer is option A, \(\frac{8}{7}\). Treat the division sign carefully: the expression is the first fraction divided by the second. First, \(\frac{16!}{14!\cdot2!}=\frac{16\times15}{2}=120\), because \(16!/14!=16\times15\). Second, \(\frac{15!}{13!\cdot2!}=\frac{15 imes14}{2}=105\). Therefore the required value is \(120\div105=\frac{120}{105}=\frac{8}{7}\), after dividing by 15. Option A is correct. Option B, \(\frac{16}{15}\), can arise from cancelling incorrectly and does not equal 120 divided by 105. Option C, \(\frac{15}{14}\), is not the ratio of the two evaluated fractions. Option D, \(\frac{4}{3}\), is another incorrect cancellation. The memory cue is: simplify each fraction separately, then perform the indicated division.

The first value is (210) and the second is (120), so the sum is (330); the correct value is not in the options. In such a case, the question or options are incorrect.

(11\cdot10\cdot9=990) and (10\cdot9\cdot8=720), so the difference is (270). Expand three consecutive factors.

On cancelling \((n+3)!\) from \(\frac{(n+5)!}{(n+3)!}\), we get \((n+5)(n+4)=132\). The consecutive factors of 132 are \(11\) and \(12\), so \(n+4=11\) and hence \(n=7\). For instance, if \(n=6\), the product is \(11\times10=110\), so it is not correct. Exam tip: In factorial quotients, expand the larger factorial only up to the smaller factorial and cancel common terms.

First simplify each factorial quotient separately. For the first term, \(\frac{20!}{18!\,2!}=\frac{20\times19\times18!}{18!\times2}=\frac{20\times19}{2}=190\). For the second, \(\frac{19!}{17!\,2!}=\frac{19\times18\times17!}{17!\times2}=\frac{19\times18}{2}=171\). The common factorial parts cancel before any subtraction is done.

The question asks for the difference, so subtract the second value from the first: \(190-171=19\). Equivalently, the expressions are consecutive combination values \(\binom{20}{2}\) and \(\binom{19}{2}\), whose difference is 19. Therefore option B is correct. Options 18 and 20 can result from subtracting nearby factors incorrectly, while 21 does not follow from the calculation.

( \frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot2\cdot1}=495 ). Reduce the large product step by step.

The two terms are (126) and (84), so the sum is (210); the correct option should be (210). Always verify the option after calculation.

\(\frac{8!}{5!}=8\times7\times6=336\), since \(5!\) cancels. Now \(3!=6\), so \(336\div6=56\). Option 336 is only the value of \(\frac{8!}{5!}\); it does not include division by \(3!\). Exam tip: expand factorials and cancel common factorial terms first.

( \frac{6!}{3!}=120 ) and ( \frac{4!}{5!}=\frac{1}{5} ), so the value is (24). Be careful with reversed factorial ratios.