Factorial notation

Write \((n+3)!=(n+3)(n+2)(n+1)n(n-1)!\). Cancelling \((n-1)!\), \((n+2)\), and \((n+1)\) gives \(n(n+3)=108\). For \(n=9\), \(9\times12=108\), so the correct answer is 9. For instance, \(n=8\) gives \(8\times11=88\), not 108. Exam tip: In factorial problems, expand the larger factorial up to the smaller one before cancelling common factors.

The exponent is ( \left\lfloor\frac{25}{3}\right\rfloor+\left\lfloor\frac{25}{9}\right\rfloor=10 ). Adding the contribution of (9) is necessary.

Direct answer: Option B, n=9. We need the smallest n! containing at least \\(2^7\cdot3^4\cdot5\\). For 9!, count prime factors: the factors 2, 4 and 8 contribute \\(1+2+3=6\\) twos, and 6 contributes one more, giving seven twos. The factors 3 and 6 contribute two threes, and 9 contributes two more, giving four threes. The factor 5 supplies one five. Thus 9! is divisible by the required product. For 8!, there are only six factors of 2 and three factors of 3, so it fails. Therefore every smaller factorial also fails, making n=9 smallest. Option A, 8, is insufficient. Option B works. Options C, 10, and D, 11, also work but are not smallest. Exam cue: count prime powers in n!; do not merely check whether each prime appears once.

\(\frac{25!}{23!\cdot2!}=\frac{25\times24}{2}=300\) and \(\frac{24!}{22!\cdot2!}=\frac{24\times23}{2}=276\). Therefore, the difference is \(300-276=24\). Option 23 is only a factor occurring while simplifying the second term; it is not the required difference. Exam tip: In factorial fractions, expand the larger factorial only up to the common factorial, then cancel it first.

\(\frac{16!}{12!\cdot4!}=\binom{16}{4}=1820\), while \(\frac{15!}{12!\cdot3!}=\binom{15}{3}=455\). Therefore, the required value is \(1820-455=1365\). Option 1450 is incorrect because the difference of the two combination terms is 1365. Exam tip: Recognising factorial expressions as \(\binom{n}{r}\) often makes calculations quicker.

\((n+1)!=(n+1)n(n-1)(n-2)(n-3)!\). On dividing by \((n-3)!\), the common factorial cancels, leaving \((n+1)n(n-1)(n-2)\). Option B incorrectly omits the factor \((n-2)\). Exam tip: In a factorial quotient, expand the numerator down to the denominator factorial and then cancel common terms.

Cancelling \((n-3)!\) from \((n+1)!\) gives \(\frac{(n+1)!}{(n-3)!}=(n+1)n(n-1)(n-2)\). Hence, \((n+1)n(n-1)(n-2)=5040\). For \(n=9\), \(10\times9\times8\times7=5040\), so the correct value is \(9\). For \(n=8\), the product is \(9\times8\times7\times6=3024\), not 5040. Exam tip: In factorial ratios, expand the larger factorial only up to the smaller factorial and then cancel common terms.

\((n+2)!/(n-2)!=(n+2)(n+1)n(n-1)\), while \((n+1)!/(n-3)!=(n+1)n(n-1)(n-2)\). Cancelling the common factors \((n+1)n(n-1)\) gives \(\frac{n+2}{n-2}\). Option B uses incorrect remaining factors. Exam tip: In factorial quotients, expand the larger factorial only up to the smaller one, then cancel common factors.

The given expression can be written as \(\frac{(n+2)!}{(n-2)!}\times\frac{(n-3)!}{(n+1)!}\). Cancelling common factorial factors gives \(\frac{n+2}{n-2}\). Hence, \(\frac{n+2}{n-2}=2\Rightarrow n+2=2n-4\Rightarrow n=6\). Therefore, 6 is correct. If 5 is substituted, the value is \(\frac{7}{3}\), not 2. Exam tip: In factorial fractions, expand only enough consecutive factors to cancel common terms.

The number of zeros is ( \left\lfloor\frac{30}{5}\right\rfloor+\left\lfloor\frac{30}{25}\right\rfloor=7 ). Add the extra contribution of (25).

Direct answer: Option B, \\(21/5\\). First simplify each factorial ratio: \\(22!/20!=22\cdot21\\), because all factors through 20 cancel. Similarly, \\(11!/9!=11\cdot10\\). The division therefore becomes \\( (22\cdot21)/(11\cdot10)\\). Cancel 22 with 11 to get 2, so the result is \\(2\cdot21/10=42/10=21/5\\). Option A, 19/5, does not follow from the uncancelled factors. Option B matches the exact calculation. Option C, 23/5, uses incorrect neighbouring factors. Option D, 25/5, is also based on incorrect expansion and equals 5, not the required value. The safe rule is to expand a factorial ratio only by the number of factors missing from the denominator, then cancel before multiplying. Memory cue: \\(m!/(m-2)!=m(m-1)\\).

Write \(7!=7\cdot6!\) and \(9!=9\cdot8!\). Then
\[
\frac{7!\cdot9!}{8!\cdot6!}=\frac{(7\cdot6!)(9\cdot8!)}{8!\cdot6!}=7\cdot9=63.
\]
Therefore, the correct answer is 63. A value such as 72 may result from cancelling consecutive factorial terms incorrectly. Exam tip: use \(\frac{n!}{(n-1)!}=n\) to simplify factorial fractions quickly.

Write \((n+4)!=(n+4)(n+3)!\). The numerator becomes \((n+3)![(n+4)-1]=(n+3)!(n+3)\). Since \((n+3)!=(n+3)(n+2)!\), we get \(\frac{(n+3)!(n+3)}{(n+2)!}=(n+3)^2\). Hence, option A is correct. \((n+4)^2\) incorrectly retains an extra factor of \(n+4\). Exam tip: In factorial expressions, first factor out the smallest common factorial.

Using \((n+4)!=(n+4)(n+3)!\), the numerator becomes \((n+3)![(n+4)-1]=(n+3)!(n+3)\). Hence the expression is \(\frac{(n+3)!(n+3)}{(n+2)!}=(n+3)^2\). Therefore, \((n+3)^2=144\). Since \(n\) must be a non-negative integer for the factorials here, \(n+3=12\), giving \(n=9\). The root \(n+3=-12\) gives \(n=-15\), which is not valid. Exam tip: In factorial questions, first expand the larger factorial in terms of the smaller one and cancel common factors.

\(\frac{(n+5)!}{(n+2)!}=(n+5)(n+4)(n+3)\), while \(\frac{(n+4)!}{(n+1)!}=(n+4)(n+3)(n+2)\). Taking \((n+4)(n+3)\) as the common factor gives \((n+4)(n+3)[(n+5)-(n+2)] = 3(n+4)(n+3)\). Hence, option B is correct. Option D misses the factor 3. Exam tip: In factorial quotients, expand only up to the denominator factorial and cancel first.

\(\frac{(n+5)!}{(n+2)!}=(n+5)(n+4)(n+3)\) and \(\frac{(n+4)!}{(n+1)!}=(n+4)(n+3)(n+2)\). Their difference is \((n+4)(n+3)[(n+5)-(n+2)]=3(n+4)(n+3)\). Hence, \((n+4)(n+3)=132=12\times11\), giving \(n=8\). For option 7, the product would be \(11\times10\), not 132. Exam tip: In factorial fractions, first cancel the common factorial terms and write the remaining consecutive factors.

In the given expression, \(4!\) cancels out. Next, \(\frac{18!}{17!}=18\) and \(\frac{13!}{14!}=\frac{1}{14}\). Hence, the value is \(18\times\frac{1}{14}=\frac{9}{7}\). \(\frac{153}{119}\) is not fully simplified; it also reduces to \(\frac{9}{7}\). Exam tip: In factorial divisions, cancel common factorial terms first and then simplify consecutive factors.

Since \((n+2)!=(n+2)(n+1)n!\), \(\frac{(n+2)!}{n!}=(n+2)(n+1)=n^2+3n+2\). Also, \(n!=n(n-1)(n-2)!\), so \(\frac{n!}{(n-2)!}=n(n-1)=n^2-n\). Adding gives \(n^2+3n+2+n^2-n=2n^2+2n+2\). Option A misses the constant term \(2\). Exam tip: in factorial quotients, cancel the common factorial first and then expand the remaining consecutive factors.

\(\frac{(n+2)!}{n!}=(n+2)(n+1)\) and \(\frac{n!}{(n-2)!}=n(n-1)\). Thus, \((n+2)(n+1)+n(n-1)=114\). Simplifying gives \(2n^2+2n+2=114\), or \(n^2+n-56=0\). Hence, \((n+8)(n-7)=0\), so \(n=7\) or \(-8\). Since the factorial expression requires \(n\ge2\), the valid value is \(n=7\). Substituting \(n=8\) gives 146 on the left-hand side, so it is a close but incorrect option. Exam tip: cancel factorials first by expanding only the required consecutive factors.

The first term is \(\frac{12!}{6!6!}=\binom{12}{6}=924\). The second term is \(\frac{11!}{5!6!}=\binom{11}{5}=462\). Therefore, the required value is \(924-462=462\). An answer such as 420 can result from incomplete simplification of the factorial terms. Exam tip: Recognise expressions of the form \(\frac{n!}{r!(n-r)!}\) as combinations to calculate quickly.

\(\frac{(n+6)!}{(n+2)!}=(n+3)(n+4)(n+5)(n+6)\) and \(\frac{(n+5)!}{(n+1)!}=(n+2)(n+3)(n+4)(n+5)\). Taking the common factors, their difference becomes \(4(n+3)(n+4)(n+5)\). Hence \(4(n+3)(n+4)(n+5)=2880\), so \((n+3)(n+4)(n+5)=720\). Since \(8\times9\times10=720\), \(n=5\). For the nearby option 6, the product would be \(9\times10\times11\), not 720. Exam tip: first expand factorial ratios into consecutive factors and cancel common terms carefully.

(10!) contains exponent (8) of (2) and exponent (2) of (5). (9!) does not contain (5^2), so (10) is minimum.

The exponent is ( \left\lfloor\frac{100}{7}\right\rfloor+\left\lfloor\frac{100}{49}\right\rfloor=16 ). Always add the contribution of higher powers such as (49).

\(\frac{19!}{16!}=17\times18\times19\) and \(\frac{18!}{15!}=16\times17\times18\). Therefore, the expression becomes \(17\times18\times(19-16)=17\times18\times3=918\). Hence, 918 is correct. A value such as 1026 may result from incorrect cancellation or subtraction. Exam tip: In factorial fractions, cancel the smaller factorial first, then factor out common consecutive terms where possible.

Write \(16! = 16\cdot15\cdot14\cdot13\cdot12!\), so \(12!\) cancels. Hence, \(\frac{16!}{12!\cdot4!}=\frac{16\cdot15\cdot14\cdot13}{4\cdot3\cdot2\cdot1}=1820\). Therefore, option B is correct. This expression is \(\binom{16}{4}\); 1365 is the value of \(\binom{15}{4}\), not of the given expression. Exam tip: In factorial fractions, cancel common factorial terms before carrying out the remaining multiplication.