Factorial notation

Expand each factorial quotient only as far as necessary. Since \\(18!=18\cdot17\cdot16\cdot15!\\), we get \\(18!/15!=18\cdot17\cdot16=4896\\). Similarly, \\(17!=17\cdot16\cdot15\cdot14!\\), so \\(17!/14!=17\cdot16\cdot15=4080\\). The required difference is therefore \\(4896-4080=816\\), making option B correct.

It is important not to calculate the complete factorials, because the common lower factorials cancel directly. Both products contain \\(17\cdot16\\), so the difference can also be checked as \\(17\cdot16(18-15)=272\cdot3=816\\). Thus the given computation and answer are consistent.

Cancelling \(11!\) from \(15!\) gives \(\frac{15!}{11!\cdot4!}=\frac{15\times14\times13\times12}{4\times3\times2\times1}\). Simplifying this expression gives \(1365\). The value \(1820\) is \(\binom{16}{4}\), so it does not apply here. Exam tip: use \(\frac{n!}{r!(n-r)!}=\binom{n}{r}\); here the expression is \(\binom{15}{4}\).

\(\frac{(n+4)!}{n!}=(n+4)(n+3)(n+2)(n+1)\). Thus, the product of four consecutive integers must be \(11880\). Since \(9\times10\times11\times12=11880\), we get \(n+1=9\), so \(n=8\). For instance, if \(n=7\), the product is \(8\times9\times10\times11=7920\), not \(11880\). Exam tip: cancel factorials first and rewrite the expression as a product of consecutive integers.

Using \((n+2)!=(n+2)(n+1)!\), the expression becomes \(\frac{(n+2)(n+1)!+(n+1)!}{(n+1)!}\). Dividing by \((n+1)!\) gives \(n+2+1=n+3\). Hence, \(n+3=14\), so \(n=11\). If 12 were used, the expression would equal 15, not 14. Exam tip: In factorial questions, rewrite the larger factorial in terms of the smaller one before simplifying.

Using \((n+3)!=(n+3)(n+2)!\), the numerator becomes \((n+2)![(n+3)-1]=(n+2)!(n+2)\). Since \((n+2)!=(n+2)(n+1)!\), \(\frac{(n+2)!(n+2)}{(n+1)!}=(n+2)^2\). The option \((n+3)^2\) does not correctly account for subtracting \((n+2)!\). Exam tip: in factorial subtraction, first factor out the smaller common factorial.

Using \((n+3)!=(n+3)(n+2)!\), the numerator becomes \((n+2)![(n+3)-1]=(n+2)(n+2)!\). Hence the given expression simplifies to \(\frac{(n+2)(n+2)!}{(n+1)!}=(n+2)^2\). Thus, \((n+2)^2=121\), giving \(n+2=11\) and \(n=9\). The other algebraic value, \(n+2=-11\), gives \(n=-13\), for which the factorials in the question are not defined. Exam tip: In factorial expressions, first expand the larger factorial in terms of the smaller one.

The simplified form is ( \frac{n(n-1)}{(n+1)(n+2)} ). Putting (n=5) gives ( \frac{20}{42}=\frac{10}{21} ).

The exponent is ( \left\lfloor\frac{20}{2}\right\rfloor+\left\lfloor\frac{20}{4}\right\rfloor+\left\lfloor\frac{20}{8}\right\rfloor+\left\lfloor\frac{20}{16}\right\rfloor=18 ). Add all quotients for a prime exponent.

The number of zeros is ( \left\lfloor\frac{45}{5}\right\rfloor+\left\lfloor\frac{45}{25}\right\rfloor=10 ). Do not forget the contribution of (25).

(1440=2^5\cdot3^2\cdot5), and this requirement is first satisfied by (8!). For divisibility, check prime factors.

\(\frac{20!}{18!\cdot2!}=\frac{20\times19}{2}=190\), and \(\frac{19!}{17!\cdot2!}=\frac{19\times18}{2}=171\). Therefore, the sum is \(190+171=361\). A value such as 371 can result from incorrectly using 19 instead of 18 in the second term. Exam tip: Expand a factorial only up to the factors needed before cancelling.

The expression is \(\binom{21}{3}-\binom{20}{3}\). Using Pascal’s identity, \(\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}\), it becomes \(\binom{20}{2}=\frac{20\times19}{2}=190\). Hence, option C is correct. The nearby option 180 is incorrect because the required difference is \(\binom{20}{2}\). Exam tip: use Pascal’s identity for differences of consecutive combinations.

Convert the division into multiplication by the reciprocal:
\(\frac{14!}{10!\cdot4!}\times\frac{9!\cdot4!}{13!}\). Cancelling \(4!\), and then using \(14!=14\cdot13!\) and \(10!=10\cdot9!\), gives \(\frac{14}{10}=\frac{7}{5}\). Hence, option B is correct. A value such as \(\frac{5}{4}\) can result from cancelling factorial terms in the wrong order. Exam tip: first change division into multiplication by the reciprocal, then cancel common factorial factors.

Using \((2n)!=(2n)(2n-1)(2n-2)(2n-3)(2n-4)!\), the factor \((2n-4)!\) cancels from the numerator and denominator. Hence the result is \((2n)(2n-1)(2n-2)(2n-3)\). Option C incorrectly omits \(2n\) and includes \(2n-4\). Exam tip: in \(\frac{m!}{(m-r)!}\), write the first \(r\) consecutive descending factors from \(m\).

Expanding the factorial ratio gives \(\frac{(2n)!}{(2n-4)!}=(2n)(2n-1)(2n-2)(2n-3)\). The number 1680 can be written as the product of four consecutive integers: \(5\times6\times7\times8\). Hence \(2n=8\), so \(n=4\). For instance, if \(n=5\), the product would be \(10\times9\times8\times7\), not 1680. Exam tip: In factorial-ratio questions, cancel the common factorial part and write the remaining consecutive factors.

Write \((3n+2)!=(3n+2)(3n+1)(3n)!\). The factor \((3n)!\) cancels from the numerator and denominator, giving \(\frac{(3n+2)!}{(3n)!}=(3n+2)(3n+1)\). Option D incorrectly retains an extra factor of \(3n\). Exam tip: expand the larger factorial only up to the smaller factorial, then cancel common factors.

Since \((3n+2)!=(3n+2)(3n+1)(3n)!\), the given ratio becomes \((3n+2)(3n+1)\). Hence, \((3n+2)(3n+1)=182\). As \(182=14\times13\), we get \(3n+2=14\), so \(n=4\). Option 5 is a close distractor, but for \(n=5\), the product is \(17\times16=272\), not 182. Exam tip: in factorial ratios, cancel the common factorial first and write the consecutive remaining factors.

\(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\) and \(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\). Hence, their difference is \((n+3)[(n+4)-(n+2)]=2(n+3)\). So \(2(n+3)=24\), giving \(n+3=12\) and \(n=9\). For example, if \(n=8\), the expression equals 22, not 24. Exam tip: In factorial quotients, cancel the common factorial terms and write the remaining consecutive factors.

Since \((n+5)!=(n+5)(n+4)!\), the numerator becomes \((n+4)![(n+5)+1]=(n+4)!(n+6)\). Also, \((n+4)!=(n+4)(n+3)!\). Therefore, \(\frac{(n+4)!(n+6)}{(n+3)!}=(n+4)(n+6)\). Option B incorrectly retains the factor \(n+5\). Exam tip: in factorial expressions, first take the smaller factorial as a common factor.

Write \((n+5)!=(n+5)(n+4)!\). Then the numerator becomes \((n+4)![(n+5)+1]\), so the expression simplifies to \((n+4)(n+6)=120\). Hence \((n+5)^2=121\), giving \(n=6\) or \(n=-16\). Since \(n+3\) must be a non-negative integer for the factorial to be defined, only \(n=6\) is valid. For instance, option 5 gives a product of 99, not 120. Exam tip: in factorial equations, first express the larger factorial in terms of the smaller one and cancel common factors.

The first term is \(\frac{12!}{8!4!}=\binom{12}{4}=495\), and the second term is \(\frac{12!}{9!3!}=\binom{12}{3}=220\). Therefore, their sum is \(495+220=715\). A nearby value such as 705 can result from an addition error. Exam tip: recognising such expressions as \(\binom{n}{r}\) makes calculation faster.

\(\frac{10!}{5!\cdot5!}=\binom{10}{5}=252\). Therefore, \(252\times\frac{5}{9}=28\times5=140\). Hence, 140 is the correct answer. Although 150 may appear close, dividing 252 by 9 gives exactly 28, so the result is 140. Exam tip: Recognising factorial ratios as combinations such as \(\binom{n}{r}\) makes calculations quicker.

\(\frac{13!}{10!}=13\times12\times11=1716\), while \(\frac{12!}{10!}=12\times11=132\). Hence, the expression is \(1716-2\times132=1716-264=1452\). Option D is only the first term, without subtracting the second term. Exam tip: Cancel the common factorial first and then multiply the remaining consecutive factors.

Use \(9!=9\cdot8\cdot7!\) and \(8!=8\cdot7!\). Hence, \(\frac{9!+2\cdot8!}{7!}=9\cdot8+2\cdot8=72+16=88\). Option 84 results from an incorrect simplification of the \(2\cdot8!\) term. Exam tip: rewrite each factorial in terms of the denominator factorial before cancelling.

Write \((n+3)!=(n+3)(n+2)(n+1)n(n-1)!\). On dividing by the given denominator, \((n-1)!\), \((n+2)\), and \((n+1)\) cancel, leaving \(n(n+3)\). Option B incorrectly retains \((n+1)\) instead of \((n+3)\). Exam tip: expand a factorial only as far as needed, then cancel common factors.