Factorial notation

Correct answer: C. (26)

The exponent is (\left\lfloor\frac{30}{2}\right\rfloor+\left\lfloor\frac{30}{4}\right\rfloor+\left\lfloor\frac{30}{8}\right\rfloor+\left\lfloor\frac{30}{16}\right\rfloor=15+7+3+1=26). Add quotients for all powers.

Correct answer: B. (17)

Factorial notation allows common factors to be cancelled before doing the arithmetic. Since 18! = 18×17×16! and 17! = 17×16×15!, the factorial fractions become products of only the required consecutive integers. This makes the subtraction simple and avoids calculating the very large factorials themselves.

The first fraction is \(\frac{18!}{16!\times2!}=\frac{18\times17}{2}=153\). The second is \(\frac{17!}{15!\times2!}=\frac{17\times16}{2}=136\). Their difference is \(153-136=17\). Equivalently, factor out 17 and use \(\frac{17(18-16)}{2}=17\). Thus option B is correct.

Correct answer: C. 20

\(\frac{21!}{19!\times2!}=\frac{21\times20}{2}=210\), and \(\frac{20!}{18!\times2!}=\frac{20\times19}{2}=190\). Therefore, the required difference is \(210-190=20\). Hence, option C is correct. The nearby answer 19 may result from an error while cancelling factors. Exam tip: In factorial fractions, expand the larger factorial only up to the required factors and then cancel common terms.

Correct answer: C. 10

On cancelling \((n+3)!\) from \((n+5)!\), we get \(\frac{(n+5)!}{(n+3)!}=(n+5)(n+4)\). Thus, \((n+5)(n+4)=210\). The consecutive factors of 210 are \(14\) and \(15\), so \(n+4=14\), giving \(n=10\). For \(n=9\), the product is \(13\times14=182\), not 210. Exam tip: In factorial fractions, first cancel the common factorial part and write the remaining consecutive factors.

Correct answer: B. 441

Since \(22!=22\times21!\), \(22!-21!=21!(22-1)=21\times21!\). Also, \(21!=21\times20!\), so \(\frac{21\times21!}{20!}=21\times21=441\). The value 462 does not follow from the correct factorial simplification. Exam tip: In subtraction involving factorials, first factor out the smaller common factorial.

Correct answer: C. (18)

The exponent is (\left\lfloor\frac{75}{5}\right\rfloor+\left\lfloor\frac{75}{25}\right\rfloor=15+3=18). Add quotients of all possible powers of the prime.

Correct answer: D. (2024)

(\frac{24!}{21!\times3!}=\frac{24\times23\times22}{6}=2024). Avoid expanding the whole large factorial.

Correct answer: B. (11)

The direct answer is option B, \(n=11\). Use the factorial cancellation rule: \(\frac{n!}{(n-5)!}=n(n-1)(n-2)(n-3)(n-4)\). We need five consecutive descending factors whose product is 55440. Test option B: \(11\cdot10\cdot9\cdot8\cdot7\). Stepwise, \(11\cdot10=110\), \(9\cdot8=72\), and \(110\cdot72\cdot7=7920\cdot7=55440\). Thus \(n=11\). Option A gives \(10\cdot9\cdot8\cdot7\cdot6=30240\), not the required value. Option C gives \(12\cdot11\cdot10\cdot9\cdot8=95040\). Option D gives an even larger product, \(13\cdot12\cdot11\cdot10\cdot9=154440\). Also, factorials require \(n\ge5\), which all choices satisfy. Memory cue: cancel factorials first, then count exactly five descending factors.

Correct answer: A. 21

Since 343 = 7^3, n! must contain at least three factors of 7. In 21!, the numbers 7, 14, and 21 contribute three factors of 7, so 7^3 divides 21!. In contrast, 20! contains only two factors of 7, from 7 and 14, so it is not divisible by 343. Exam tip: For a prime power p^k, count the factors of p present in n!.

Correct answer: B. 58

\(\frac{30!}{28!}=30\times29\) because \(28!\) cancels. Similarly, \(\frac{29!}{27!}=29\times28\). Therefore, the difference is \(30\times29-29\times28=29(30-28)=58\). Option 56 may seem close, but it does not result from the correct common factor \(29\) and difference \(2\). Exam tip: In factorial fractions, cancel the smaller factorial first and write the remaining consecutive factors.

Correct answer: A. \(10\)

Since \((n+6)!=(n+6)(n+5)(n+4)!\), we get \(\frac{(n+6)!}{(n+4)!}=(n+6)(n+5)\). Thus, \((n+6)(n+5)=272\). As \(16\times17=272\), \(n+5=16\), so \(n=10\). For \(n=11\), the product would be \(17\times18\), not 272. Exam tip: In a factorial quotient, write the remaining consecutive factors before cancelling the common factorial term.

Correct answer: A. \((n+1)!=(n+1)\cdot n!\)

By the recursive definition of factorial, \((n+1)!=(n+1)\cdot n!\), since it includes every factor from 1 to \(n+1\). Option B incorrectly treats factorial as an addition. Exam tip: rewrite \((n+1)!\) as \((n+1)n!\) before comparing factorial expressions.