Factorial notation

Correct answer: C. 26

Since \(27!=27\times26\times25!\), \(\frac{27!}{25!\cdot2!}=\frac{27\times26}{2}=351\). Similarly, \(\frac{26!}{24!\cdot2!}=\frac{26\times25}{2}=325\). Hence, the value is \(351-325=26\). Although 25 occurs as a factor in the second term, it is not the difference. Exam tip: In factorial fractions, expand the larger factorial only up to the smaller factorial and cancel common factors first.

Correct answer: C. 1820

After cancelling \(13!\), \(\frac{17!}{13!\cdot4!}=\frac{17\times16\times15\times14}{4!}=2380\). Similarly, \(\frac{16!}{13!\cdot3!}=\frac{16\times15\times14}{3!}=560\). Hence, the value of the expression is \(2380-560=1820\). \(2380\) is only the first term, so the second term must still be subtracted. Exam tip: expand the larger factorial only up to the smaller factorial, then cancel common factors before calculating.

Correct answer: C. 6

Writing the numerator up to \((n-4)!\), we get \((n+2)!=(n+2)(n+1)n(n-1)(n-2)(n-3)(n-4)!\). Cancelling \((n-4)!\) leaves 6 consecutive factors: \((n+2)(n+1)n(n-1)(n-2)(n-3)\). Seven factors would incorrectly include \(n-4\), which is already cancelled by the denominator. Exam tip: in \(\frac{a!}{b!}\), when \(a>b\), the number of remaining factors is \(a-b\).

Correct answer: B. 8

Using \((n+2)!=(n+2)(n+1)n(n-1)(n-2)!\), cancel \((n-2)!\) to obtain \((n+2)(n+1)n(n-1)=5040\). Now \(7\times8\times9\times10=5040\), so the four consecutive factors are \(n-1,n,n+1,n+2\) with \(n=8\). The close distractor \(n=9\) gives \(8\times9\times10\times11\), not 5040. Exam tip: expand the larger factorial only down to the smaller factorial before cancelling.

Correct answer: A. \(\frac{n+3}{n-2}\)

Writing the complex fraction as a product gives \(\frac{(n+3)!(n-3)!}{(n-2)!(n+2)!}\). Now use \((n+3)!=(n+3)(n+2)!\) and \((n-2)!=(n-2)(n-3)!\). The factors \((n+2)!\) and \((n-3)!\) cancel, leaving \(\frac{n+3}{n-2}\). Option D is the reciprocal of the required result, so it is incorrect. Exam tip: expand a larger factorial only until a common smaller factorial appears, then cancel common factors.

Correct answer: B. (19)

The number of zeros is ( \left\lfloor\frac{80}{5}\right\rfloor+\left\lfloor\frac{80}{25}\right\rfloor=19 ). Add the contribution of powers of (5).

Correct answer: B. ( \frac{46}{11} )

The direct answer is option B, \(\frac{46}{11}\). Simplify each factorial ratio before dividing. Since \(24!=24\cdot23\cdot22!\), \(\frac{24!}{22!}=24\cdot23\). Similarly, \(12!=12\cdot11\cdot10!\), so \(\frac{12!}{10!}=12\cdot11\). Therefore the given division is \(\frac{24\cdot23}{12\cdot11}\). Cancel 24 with 12 to get \(\frac{2\cdot23}{11}=\frac{46}{11}\). Option A, \(\frac{38}{11}\), uses an incorrect numerator; option B matches the calculation. Option C, \(\frac{52}{11}\), and option D, \(\frac{58}{11}\), also do not follow from the factorial products. The key is that division by a factorial cancels all factors up to that factorial, not just one factor.

Correct answer: C. 88

Substitute \(8!=8\cdot7!\) and \(11!=11\cdot10!\):
\(\frac{8!\cdot11!}{10!\cdot7!}=\frac{(8\cdot7!)(11\cdot10!)}{10!\cdot7!}=8\cdot11=88\). Therefore, option C is correct. After cancelling \(10!\) and \(7!\), only 8 and 11 remain; 80 is not their product. Exam tip: in factorial fractions, cancel common factorial terms before expanding or multiplying.

Correct answer: A. \((n+5)^2\)

Using \((n+6)!=(n+6)(n+5)!\), the numerator becomes \((n+5)![(n+6)-1]=(n+5)!(n+5)\). Also, \((n+5)!=(n+5)(n+4)!\). Therefore, \(\frac{(n+5)!(n+5)}{(n+4)!}=(n+5)^2\), so option A is correct. Option D incorrectly retains the factor \((n+6)\); after subtraction, the factor is \((n+6)-1=(n+5)\). Exam tip: In a factorial subtraction, factor out the smaller common factorial first.

Correct answer: B. 9

Using \((n+6)!=(n+6)(n+5)!\), the numerator becomes \((n+5)![(n+6)-1]=(n+5)(n+5)!\). Thus, the given expression equals \((n+5)^2\). Hence \((n+5)^2=196=14^2\). Since \(n+4\) must be a non-negative integer for the factorials to be defined, \(n+5=14\), not \(-14\). Therefore, \(n=9\). Substituting 8 would give \(169\), not 196. Exam tip: In factorial subtraction, factor out the smaller common factorial first.

Correct answer: B. (4(n+6)(n+5))

Take common ((n+6)(n+5)). The difference is ((n+7)-(n+4)=3), so the correct form is (3(n+6)(n+5)).

Correct answer: B. 10

\(\frac{(n+7)!}{(n+4)!}=(n+7)(n+6)(n+5)\), while \(\frac{(n+6)!}{(n+3)!}=(n+6)(n+5)(n+4)\). Factoring out the common terms gives the difference as \((n+6)(n+5)[(n+7)-(n+4)]=3(n+6)(n+5)\). Thus, \(3(n+6)(n+5)=720\), so \((n+6)(n+5)=240\). The consecutive numbers 15 and 16 have product 240, hence \(n+5=15\) and \(n=10\). For option 9, the product would be \(14\times15=210\), so it is not correct. Exam tip: Expand factorial ratios into consecutive factors before simplifying common terms.

Correct answer: B. 13

Using \(n!=n(n-1)!\) and \((n+1)!=(n+1)n!\), the given expression reduces to \(\frac{n}{n+1}\). Therefore, \(\frac{n}{n+1}=\frac{13}{14}\). Cross-multiplication gives \(14n=13n+13\), hence \(n=13\). If \(n=14\), the fraction would be \(\frac{14}{15}\), not the given value. Exam tip: expand consecutive factorials first and cancel common factors before solving.

Correct answer: B. ( \frac{19}{15} )

The ratio of the two terms directly becomes ( \frac{19}{15} ). Cancel factorials instead of finding large values.

Correct answer: A. \(2n^3+6n^2+10n+6\)

\(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)=n^3+6n^2+11n+6\). Also, \(\frac{(n+1)!}{(n-2)!}=(n+1)n(n-1)=n^3-n\). Their sum is \(2n^3+6n^2+10n+6\), so option A is correct. In option B, the coefficient of the linear term is incorrect. Exam tip: In factorial fractions, cancel the common factorial first and then write the remaining consecutive factors.

Correct answer: A. (6)

Putting (n=6) gives (9\cdot8\cdot7+7\cdot6\cdot5=504+210=714). Therefore the equation does not match the options.

Correct answer: A. 792

The first term is \(\frac{13!}{6!\cdot7!}=\binom{13}{6}=1716\), and the second term is \(\frac{12!}{6!\cdot6!}=\binom{12}{6}=924\). Hence, the value is \(1716-924=792\). Option 846 may result from evaluating the second binomial coefficient incorrectly. Exam tip: recognise \(\frac{n!}{r!(n-r)!}\) as \(\binom{n}{r}\) to calculate quickly.

Correct answer: A. (4)

After cancellation we get (n(n-1)(n+1)=120) and (n=4) satisfies it. In exams write consecutive factors first.

Correct answer: B. 6

Since \((n+2)!=(n+2)(n+1)n!\), we have \(\frac{(n+2)!}{n!}=(n+2)(n+1)\). Therefore, \((n+2)(n+1)=56\). The consecutive factors of 56 are \(7\) and \(8\), so \(n+1=7\) and hence \(n=6\). For the close distractor 7, the product is \(9\times8=72\), not 56. Exam tip: In a factorial fraction, expand the larger factorial only up to the smaller factorial and cancel common terms.

Correct answer: A. (128)

Here (\frac{9!}{7!}=72) and (\frac{8!}{6!}=56) so the sum is (128). In exams cancel larger factorials quickly.

Correct answer: C. (66)

(\frac{12!}{10!\times 2!}=\frac{12\times11}{2}=66). Do not expand the whole product in such questions.

Correct answer: C. (70)

The direct answer is option C, 70. Expand only the part of 15! not included in 13!: \(15!=15 imes14 imes13!\). Hence \(\frac{15!}{13! imes3}=\frac{15 imes14 imes13!}{13! imes3}=\frac{15 imes14}{3}=5 imes14=70\). Option A, 60, is wrong because the correct numerator factors are 15 and 14, not a product giving 60. Option B, 65, does not result from the required multiplication and division. Option C, 70, is correct. Option D, 75, is also not obtained after cancelling \(13!\) and dividing by 3. The denominator means \(13!\) multiplied by 3, so cancel \(13!\) first and then divide the remaining product by 3. Avoid expanding all factorials; it creates large numbers and more opportunities for mistakes. Memory cue: \(\frac{n!}{(n-2)!}=n(n-1)\).

Correct answer: C. 7

Since \(n!=n(n-1)(n-2)(n-3)!\), we get \(\frac{n!}{(n-3)!}=n(n-1)(n-2)\). Thus, \(n(n-1)(n-2)=210\). As \(7\times6\times5=210\), the value of \(n\) is 7. For the closest distractor, \(n=6\) gives \(6\times5\times4=120\), not 210. Exam tip: In a factorial ratio, expand the larger factorial only up to the smaller factorial and cancel common factors.

Correct answer: A. \(\frac{n!}{r!}\)

Since \(n!=1\cdot2\cdots r\cdot(r+1)(r+2)\cdots n\) and \(r!=1\cdot2\cdots r\), dividing \(n!\) by \(r!\) cancels the initial factors. Hence \((r+1)(r+2)\cdots n=\frac{n!}{r!}\). \((n-r)!\) is the factorial of the number of terms, not the product of the given terms. Exam tip: cancel the common initial factors in a factorial fraction to verify the result.

Correct answer: B. 8

Using \((n+3)!=(n+3)(n+2)(n+1)!\),
\[\frac{(n+3)!}{(n+1)!}=(n+3)(n+2).\]
Hence, \((n+3)(n+2)=110\). Since \(10\times11=110\), we get \(n+2=10\), so \(n=8\). The other algebraic root is \(n=-13\), but then \((n+1)!\) is not defined. Exam tip: In a factorial quotient, cancel the common factorial part first and write the remaining consecutive factors.