Factorial notation

The direct answer is D, \(n=5\). Expand the numerator until the denominator appears: \((2n+2)!=(2n+2)(2n+1)(2n)!\). Thus \(\frac{(2n+2)!}{(2n)!}=(2n+2)(2n+1)=132\). Among the options, if \(n=5\), the factors are \(2(5)+2=12\) and \(2(5)+1=11\), and \(12\times11=132\). Therefore D is correct. Option A, 3, gives \(8\times7=56\), not 132. Option B, 4, gives \(10\times9=90\), not 132. Option C, 6, gives \(14\times13=182\), which is too large. Option D, 5, matches exactly. The key is to cancel the common \((2n)!\) first and then solve the product of the two consecutive factors.

To simplify a factorial fraction, expand the larger factorial only as far as needed to expose the factorial in the denominator. This avoids unnecessary multiplication. The denominator contains \((m-2)!\) and the factor \((m-1)\), while the numerator contains both of these factors and one additional factor \(m\). After cancellation, only \(m\) remains, so option A is correct.

Use \(m!=m(m-1)(m-2)!\). Then \(\frac{m!}{(m-2)!(m-1)}=\frac{m(m-1)(m-2)!}{(m-2)!(m-1)}=m\), provided the factorial expression is defined and the denominator is nonzero. The factors \((m-1)\) and \((m-2)!\) cancel. Thus the simplified form is \(m\), not \(m-1\) or \(m(m-1)\).

Since \(n! = n(n-1)(n-2)!\), \(\frac{n!}{(n-2)!}=n(n-1)\). Hence, \(n(n-1)=9n\). For the factorial expression, \(n\geq2\), so dividing by \(n\) gives \(n-1=9\). Therefore, \(n=10\). Option 9 is not correct because it gives \(9\times8=72\) on the left, whereas the right side is \(81\). Exam tip: In factorial quotients, first expand the larger factorial up to the smaller factorial.

The first term is (165) and the second is (120), so the ratio is ( \frac{11}{8} ). Treat division as a ratio of fractions.

Each fraction should be simplified separately before subtracting. A factorial quotient of the form \(\frac{n!}{r!(n-r)!}\) represents a binomial coefficient, but direct cancellation is also clear here. The first term gives 495 and the second gives 330. Their difference is therefore 165, which is option A.

For the first term, \(\frac{12!}{8!4!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495\). For the second, \(\frac{11!}{7!4!}=\frac{11\times10\times9\times8}{4\times3\times2\times1}=330\). Subtracting gives \(495-330=165\). Hence option A is correct. Subtracting factorial symbols before evaluating would be an invalid method.

Using \((n+1)!=(n+1)n!\), the numerator becomes \((n+1)n!+n!\). Taking \(n!\) common gives \((n+2)n!\). On dividing by \(n!\), the result is \(n+2\). The expression \(n+1\) accounts only for \((n+1)!/n!\); the additional term \(n!/n!=1\) must also be included. Exam tip: In factorial expressions, first rewrite the larger factorial as a multiple of the smaller one.

Since \((n+1)!=(n+1)n!\), \(\frac{(n+1)!+n!}{n!}=\frac{(n+1)n!+n!}{n!}=n+2\). Hence, \(n+2=13\), giving \(n=11\). If \(n=12\), the left-hand side becomes 14, so it is not correct. Exam tip: In factorial expressions, first rewrite the larger factorial as a multiple of the smaller factorial.

Since \(15! = 15\cdot14\cdot13!\), \(\frac{15!}{13!}=15\cdot14=210\). Similarly, \(14! = 14\cdot13\cdot12!\), so \(\frac{14!}{12!}=14\cdot13=182\). Hence, the required value is \(210-182=28\). A value such as 32 may result from cancelling factorial terms incorrectly. Exam tip: expand the larger factorial only up to the smaller factorial before cancelling common factors.

Factorial notation means multiplying all positive integers down to 1. For example, \(6!=6\times5\times4\times3\times2\times1\). In a sum of fractions, calculate each fraction separately before adding the results. Cancelling factorial factors is often the quickest method.

The first term is \(\frac{6!}{3!}=\frac{720}{6}=120\). The second term is \(\frac{5!}{2!}=\frac{120}{2}=60\). Adding them gives \(120+60=180\). Therefore option D is correct. A common mistake is to add or cancel factorial symbols across the plus sign; cancellation can be done within each fraction, but the two separate terms must then be added normally.

\(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\) and \(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\). Hence, their sum is \((n+3)[(n+4)+(n+2)]=2(n+3)^2\). Thus \(2(n+3)^2=200\), giving \((n+3)^2=100\). For the valid non-negative integer factorial domain, \(n+3=10\), so \(n=7\). The other algebraic value, \(n=-13\), does not make the factorial terms valid. Exam tip: in factorial quotients, cancel common factors and write the remaining consecutive factors.

Using \((n+3)!=(n+3)(n+2)(n+1)n!\), we get \(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)\). Dividing this by \(n+3\) leaves \((n+2)(n+1)\). Option B is not correct because it still contains the factor \(n+3\). Exam tip: In factorial divisions, expand the larger factorial only up to the smaller factorial, then cancel common factors.

The first term is (286) and the second is (220), so the difference is (66). In such questions, write each term value and subtract.

(72=2^3\cdot3^2), and (6!) contains all these factors. For divisibility, check prime factors.

The number of ending zeros comes from multiples of (5), and (18!) has (5,10,15). Therefore, there are (3) zeros.

The exponent is ( \left\lfloor\frac{10}{2}\right\rfloor+\left\lfloor\frac{10}{4}\right\rfloor+\left\lfloor\frac{10}{8}\right\rfloor=8 ). For prime exponents, add the quotients.

The exponent is ( \left\lfloor\frac{25}{5}\right\rfloor+\left\lfloor\frac{25}{25}\right\rfloor=6 ). Do not forget higher multiples such as (25).

( \frac{n!}{(n-1)!}=n ) and ( \frac{n!}{(n+1)!}=\frac{1}{n+1} ), so the answer is ( \frac{n}{n+1} ). Break the ratio into two smaller parts.

Use the two adjacent-factorial identities (n−1)! = n!/n and (n+1)! = (n+1)n!. The denominator becomes [(n!/n)((n+1)n!)] = ((n+1)/n)(n!)^2. Consequently, the given fraction simplifies to n/(n+1). Equating this to 5/6 gives n/(n+1)=5/6. Cross-multiplication yields 6n = 5n+5, so n=5. Substitution confirms that 5!^2/(4!·6!) = 5/6, so option A is correct. The other choices would give 6/7, 7/8, and 8/9 respectively, none of which equals 5/6. The key step is simplifying before substituting numerical values.

Cancelling the common factorial terms gives \(\frac{(n+2)!}{(n-2)!}=(n+2)(n+1)n(n-1)\). Thus, the product of four consecutive integers must be 1680. Since \(8\times7\times6\times5=1680\), we get \(n-1=5\), so \(n=6\). For the closest distractor, \(n=7\) gives \(9\times8\times7\times6\), not 1680. Exam tip: In factorial ratios, expand the larger factorial first and cancel the common factorial part.

The value is ( \frac{18\cdot17\cdot16}{6\cdot5\cdot4}=\frac{204}{5} ). In division, convert both sides into products of equal length.

The direct answer is option A: \(3n(3n-1)\). Factorial means multiplying all positive integers down to 1. Therefore \((3n)!=(3n)(3n-1)(3n-2)\cdots1\). Also, \((3n-2)!=(3n-2)(3n-3)\cdots1\). When dividing, every factor from \((3n-2)\) down to 1 cancels, leaving \(3n(3n-1)\). Option A is exactly this result. Option B incorrectly uses \(3n-2\) instead of \(3n-1\) as the second remaining factor. Option C leaves factors that belong to the denominator and misses the factor \(3n\). Option D treats the result as \((3n)^2\), but the two remaining consecutive factors are different, not both \(3n\). The rule works for values where the factorials are defined, ordinarily non-negative integer arguments. Memory cue: expand only the larger factorial until the smaller factorial appears, then cancel.

Cancelling \((3n-2)!\) from \((3n)!\) gives \(3n(3n-1)=132\). Since \(12\times 11=132\), we get \(3n=12\). Therefore, \(n=4\). The alternative value \(3n=-11\) is invalid because factorials are not defined for negative integers. Exam tip: In factorial ratios, expand the larger factorial only up to the factors needed for cancellation.

Write \((2n)!=(2n)(2n-1)(2n-2)(2n-3)!\). The common factor \((2n-3)!\) cancels with the denominator, leaving \((2n)(2n-1)(2n-2)\). Option B incorrectly includes \(2n-3\) instead of starting with \(2n\). Exam tip: in \(\frac{m!}{(m-r)!}\), write the first \(r\) consecutive factors descending from \(m\).

Cancelling \((2n-3)!\) from \((2n)!\) gives \(\frac{(2n)!}{(2n-3)!}=(2n)(2n-1)(2n-2)\). Hence, \((2n)(2n-1)(2n-2)=720\). Since \(10\times9\times8=720\), we get \(2n=10\), so \(n=5\). For option 6, the product would be \(12\times11\times10\), not 720. Exam tip: In factorial ratios, expand the larger factorial only up to the smaller factorial and then cancel common terms.

\(\frac{9!}{7!}=9\times8=72\) because \(7!\) cancels out. Similarly, \(\frac{8!}{6!}=8\times7=56\). Therefore, the required difference is \(72-56=16\). Option 20 can result from an error in multiplication or subtraction. Exam tip: In a factorial fraction, cancel the common smaller factorial first and multiply only the remaining consecutive numbers.