Factorial notation

Correct answer: B. ( \frac{n(n-1)}{(n+1)(n+2)} )

( \frac{n!}{(n-2)!}=n(n-1) ) and ( \frac{n!}{(n+2)!}=\frac{1}{(n+2)(n+1)} ), so the answer follows. Break the ratio into two parts.

Correct answer: B. 8

\(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)\) and \(\frac{(n+2)!}{(n-1)!}=n(n+2)(n+1)\). Their difference is \((n+2)(n+1)[(n+3)-n]=3(n+2)(n+1)\). Hence, \((n+2)(n+1)=90\). Since \(10\times9=90\), we get \(n+2=10\), so \(n=8\). For option 9, the product would be \(11\times10=110\), so it is not correct. Exam tip: In factorial fractions, first cancel common factorial terms and write the remaining consecutive factors.

Correct answer: B. (4)

The number of ending zeros is ( \left\lfloor\frac{20}{5}\right\rfloor=4 ). To count zeros, count multiples of (5).

Correct answer: B. (7)

The exponent is ( \left\lfloor\frac{30}{5}\right\rfloor+\left\lfloor\frac{30}{25}\right\rfloor=7 ). Do not forget higher multiples such as (25).

Correct answer: B. (6)

The exponent is ( \left\lfloor\frac{15}{3}\right\rfloor+\left\lfloor\frac{15}{9}\right\rfloor=6 ). For a prime exponent, add the quotients.

Correct answer: B. (6)

(180=2^2\cdot3^2\cdot5), and (6!) contains all these factors. For divisibility, check prime factors.

Correct answer: B. (20)

The first term is (210) and the second is (190), so the difference is (20). Subtract nearby terms carefully.

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

The two terms are (495) and (330), so the ratio is ( \frac{3}{2} ). Convert large values into a simple ratio.

Correct answer: B. 546

\(\frac{8!}{5!}=8\times7\times6=336\), since \(5!\) cancels out. Similarly, \(\frac{7!}{4!}=7\times6\times5=210\). Therefore, the value is \(336+210=546\). Option 504 is only the value of neither complete sum and is incorrect. Exam tip: Expand the larger factorial only up to the smaller factorial, then cancel the common factorial term.

Correct answer: B. 56

\(\frac{9!}{4!5!}=\binom{9}{4}=126\) and \(\frac{8!}{4!4!}=\binom{8}{4}=70\). Therefore, the required value is \(126-70=56\). Option 54 is a close distractor, but it does not result from the correct subtraction. Exam tip: Recognising \(\frac{n!}{r!(n-r)!}\) as \(\binom{n}{r}\) makes such calculations quicker.

Correct answer: B. 7

Factorials are products of consecutive positive integers. When one factorial is divided by a smaller nearby factorial, all common factors cancel. Therefore (n+5)!/(n+2)!=(n+5)(n+4)(n+3). The equation becomes (n+5)(n+4)(n+3)=1320, so the answer can be checked by testing the listed values.

For option B, n=7, the three factors are 12,11,10. Their product is 12 times 11 times 10=1320, exactly the required value. The other listed values produce different products, so they do not satisfy the equation. Thus the supplied answer, option B, is correct. The factorials also require the usual nonnegative-integer interpretation of n in this school context.

Correct answer: C. 10

\(\frac{(n+2)!}{(n-1)!}=n(n+1)(n+2)\) and \(\frac{(n+1)!}{(n-2)!}=(n-1)n(n+1)\). Hence, their difference is \(n(n+1)[(n+2)-(n-1)]=3n(n+1)\). Thus, \(3n(n+1)=330\), so \(n(n+1)=110\). Since \(10\times11=110\), \(n=10\). The nearby option 11 is incorrect because \(11\times12\neq110\). Exam tip: In factorial fractions, cancel common factorial terms and write the remaining consecutive factors.

Correct answer: B. 18

Using \(19!=19\times18\times17!\) and \(2!=2\),
\[
\frac{19!}{17!\cdot2!}\times\frac{2}{19}
=\frac{19\times18\times17!}{17!\times2}\times\frac{2}{19}=18.
\]
Therefore, the correct value is 18. The factors 19, \(17!\), and 2 cancel out. Exam tip: In factorial expressions, expand the larger factorial only up to the smaller factorial before cancelling common factors.

Correct answer: B. 36

Since \(6!=6\times5\times4!=30\times4!\), \(5!=5\times4!\), and \(4!=1\times4!\), we get \(\frac{6!+5!+4!}{4!}=30+5+1=36\). Hence, option B is correct. The value 35 results if the term \(4!/4!=1\) is mistakenly omitted. Exam tip: factor out the common factorial before calculating individual factorial values.

Correct answer: D. 18

\(\frac{10!}{8!}=10\times9=90\), since \(8!\) cancels out. Similarly, \(\frac{9!}{7!}=9\times8=72\). Therefore, the required difference is \(90-72=18\). Option 16 may result from an error in multiplication or subtraction. Exam tip: expand factorials only up to the smaller factorial and cancel the common part.

Correct answer: D. (n)

The direct answer is option D, \(n\). Factorials are products of consecutive positive integers: \((n+4)!=(n+4)(n+3)(n+2)(n+1)n!\). Therefore, \(\frac{(n+4)!}{n!}=\frac{(n+4)(n+3)(n+2)(n+1)n!}{n!}=(n+4)(n+3)(n+2)(n+1)\). The factor \(n!\) cancels completely, and the expansion stops at \(n+1\), so \(n\) itself does not appear. Option A, \(n+1\), appears; option B, \(n+2\), appears; option C, \(n+3\), appears; option D, \(n\), does not appear. A common mistake is to start the product at \(n\); remember that dividing by \(n!\) removes all factors through \(n\).

Correct answer: A. (1680)

Putting (n=4) gives ( \frac{8!}{4!}=8\cdot7\cdot6\cdot5=1680 ). Understand the general form before substituting.

Correct answer: C. ( \frac{255}{28} )

The value is \( \frac{17\cdot16\cdot15}{8\cdot7\cdot6}=\frac{255}{28} \). In division, write three factors on both sides.

Correct answer: B. (715)

The first term is (1001) and the second is (286), so the difference is (715). In such questions, calculate both terms separately and subtract.

Correct answer: D. ( \frac{n+4}{n} )

After canceling the common factors in numerator and denominator, ( \frac{n+4}{n} ) remains. First convert large ratios into products.

Correct answer: C. (11)

The exponent is ( \left\lfloor\frac{14}{2}\right\rfloor+\left\lfloor\frac{14}{4}\right\rfloor+\left\lfloor\frac{14}{8}\right\rfloor=11 ). Add all quotients while finding a prime exponent.

Correct answer: B. (7)

(1008=2^4\cdot3^2\cdot7), and (7!) contains all these factors. For divisibility, satisfy all prime factor requirements.

Correct answer: A. \((n+3)(n+5)\)

Since \((n+4)!=(n+4)(n+3)!\), the numerator becomes \((n+3)![(n+4)+1]=(n+3)!(n+5)\). Using \((n+3)!=(n+3)(n+2)!\), we get \(\frac{(n+3)!(n+5)}{(n+2)!}=(n+3)(n+5)\). Option C incorrectly retains \(n+4\) instead of obtaining \(n+5\). Exam tip: when adding factorial terms, factor out the smaller factorial first.

Correct answer: B. 7

Using \((n+4)!=(n+4)(n+3)!\), the numerator becomes \((n+3)![(n+4)+1]=(n+3)!(n+5)\). Thus, the expression simplifies to \((n+3)(n+5)=120\). For \(n=7\), this gives \(10\times12=120\), so the correct value is \(7\). For instance, \(n=6\) gives \(9\times11=99\), not 120. Exam tip: In factorial fractions, first factor out the common factorial term before simplifying.

Correct answer: D. (660)

The first term is (1320) and the second is (2\cdot330=660), so the difference is (660). Complete the multiplication in the coefficient term first.