Factorial notation

Correct answer: C. 21

Using \(n!=n\times(n-1)!\), \(\frac{8!}{7!}=8\), \(\frac{7!}{6!}=7\), and \(\frac{6!}{5!}=6\). Therefore, the expression equals \(8+7+6=21\). An answer such as 20 can arise from adding terms without simplifying each factorial ratio correctly. Exam tip: directly use \(\frac{n!}{(n-1)!}=n\).

Correct answer: B. \(\frac{42}{5}\)

\(\frac{9!}{5!4!}=\binom{9}{4}=126\), while \(\frac{6!}{4!2!}=\binom{6}{2}=15\). Thus, the expression equals \(126\div15=\frac{126}{15}=\frac{42}{5}\). Therefore, option B is correct. Getting \(\frac{28}{5}\) usually results from evaluating \(\binom{9}{4}\) incorrectly or simplifying the fraction incorrectly. Exam tip: recognise \(\frac{n!}{r!(n-r)!}\) as \(\binom{n}{r}\) for faster calculation.

Correct answer: A. \((n+4)(n+3)\)

Expand the larger factorial up to the smaller factorial: \((n+4)!=(n+4)(n+3)(n+2)!\). Thus, \(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\), since \((n+2)!\) cancels. Option B incorrectly has \(n+2\) instead of \(n+3\). Exam tip: In a factorial quotient, expand the numerator factorial only until the denominator factorial appears.

Correct answer: C. 6

Using \(n!=n(n-1)(n-2)(n-3)!\), we get \(\frac{n!}{(n-3)!}=n(n-1)(n-2)\). Hence, \(n(n-1)(n-2)=120\). For \(n=6\), \(6\times5\times4=120\), so 6 is correct. The closest option, \(n=5\), gives \(5\times4\times3=60\), not 120. Exam tip: In a factorial fraction, cancel the common factorial term and write the remaining consecutive factors.

Correct answer: A. 12

Apply factorial cancellation before subtracting. Since 7! = 7 × 6 × 5!, the first quotient is 7!/5! = 7 × 6 = 42. Similarly, 6! = 6 × 5 × 4!, so 6!/4! = 6 × 5 = 30. The required value is therefore 42 − 30 = 12, making option A correct. Option D is only the second quotient and ignores the subtraction. Option C may come from an incorrect subtraction, while option B has no valid result from the two simplified terms. The expression 7!/5! must not be read as 7/5; all consecutive factors between the numerator and denominator factorials remain after cancellation. The same principle applies to the second quotient.

Correct answer: B. 24

Since \(5!=5\times4!\), we get \(4!+5!=4!(1+5)=6\times4!\). Hence, \(\frac{4!+5!}{3!}=\frac{6\times4!}{3!}=6\times4=24\). Therefore, option B is correct. The value 20 may result from incorrectly taking \(5!\) as \(4\times4!\). Exam tip: in factorial expressions, factor out the common factorial first to simplify the calculation quickly.

Correct answer: C. (56)

(\frac{12!}{10!,2!}=66) and (\frac{5!}{3!,2!}=10), so the difference is (56). Keep the larger and smaller terms separate.

Correct answer: B. 72

Since \(n!=n(n-1)(n-2)!\), \(\frac{n!}{(n-2)!}=n(n-1)\). The question gives \(n(n-1)=72\), so the value of the expression is \(72\). \(144\) could arise from incorrectly doubling the given value. Exam tip: In a factorial fraction, expand the numerator only up to the factorial in the denominator, then cancel common factors.

Correct answer: C. 120

\(\frac{6!}{5!}=6\) and \(\frac{5!}{3!}=5\times4=20\), because the common factorial terms in the numerator and denominator cancel. Hence, the expression equals \(6\times20=120\). Getting \(100\) indicates incorrect expansion or cancellation of factorial terms. Exam tip: In factorial ratios, cancel the common factorial terms first, then multiply the remaining consecutive integers.

Correct answer: C. 120

Write \(10! = 10\times 9\times 8\times 7!\), so \(7!\) cancels out. Thus, \(\frac{10!}{7!\,3!}=\frac{10\times9\times8}{3\times2\times1}=\frac{720}{6}=120\). Therefore, 120 is correct. The value 720 is only \(10\times9\times8\); division by \(3!\) is still required. Exam tip: In factorial fractions, expand the larger factorial only up to the smaller factorial and cancel first.

Correct answer: A. (7+\frac{1}{6})

(\frac{7!}{6!}=7) and (\frac{5!}{6!}=\frac{1}{6}), so the value is (7+\frac{1}{6}). Divide each term by the denominator separately.

Correct answer: B. 50

Since \(8!=8\times7\times6!\), \(a=\frac{8!}{6!}=8\times7=56\). Also, \(b=3!=3\times2\times1=6\). Hence, \(a-b=56-6=50\). Option 56 gives only the value of \(a\); it does not subtract \(b\). Exam tip: In a factorial fraction, expand the larger factorial up to the smaller one and cancel common factors first.

Correct answer: A. \((n+2)(n+1)n\)

Expanding \((n+2)!\) up to \((n-1)!\) gives \((n+2)!=(n+2)(n+1)n(n-1)!\). Cancelling \((n-1)!\) from the numerator and denominator leaves \((n+2)(n+1)n\). Option B incorrectly omits the factor \(n\). Exam tip: in factorial division, expand the larger factorial only until the smaller factorial appears, then cancel the common factorial term.

Correct answer: C. 10

For consecutive factorials, \(\frac{n!}{(n-1)!}=n\), because the common factorial terms in the numerator and denominator cancel. Hence, \(\frac{9!}{8!}=9\), \(\frac{8!}{7!}=8\), and \(\frac{7!}{6!}=7\). Therefore, the value is \(9+8-7=10\). Option 11 would result from incorrectly adding the last term instead of subtracting it. Exam tip: simplify factorial fractions by cancelling common terms first.

Correct answer: B. 4

Using \((n+1)!=(n+1)n(n-1)!\), we obtain \(\frac{(n+1)!}{(n-1)!}=n(n+1)\). Thus, \(n(n+1)=20\). Since the consecutive integers \(4\) and \(5\) have product \(20\), \(n=4\). For option \(5\), the product is \(5\times6=30\), so it is incorrect. Exam tip: In a factorial ratio, expand the larger factorial up to the smaller one and cancel common terms.

Correct answer: C. 53

\(\frac{8!}{6!}=\frac{8\times7\times6!}{6!}=8\times7=56\). Also, \(\frac{5!}{4!}=5\) and \(2!=2\). Hence, \(56-5+2=53\), so option C is correct. Option 54 can result from incorrectly evaluating \(2!\) as 3. Exam tip: In a factorial fraction, expand the larger factorial up to the smaller factorial and cancel common factors.

Correct answer: C. 42

The first term is \(\frac{7!}{4!\,3!}=\frac{7\times6\times5}{3\times2\times1}=35\). The second term is \(\frac{7!}{6!\,1!}=7\). Hence, the total is \(35+7=42\). \(35\) is only the value of the first term, while \(49\) may result from incorrectly multiplying the two terms instead of adding them. Exam tip: In factorial fractions, cancel common factorial factors before calculating.

Correct answer: B. \(3\)

Expanding \((n+3)!\) up to \(n!\) gives \((n+3)!=(n+3)(n+2)(n+1)n!\). After cancelling \(n!\), we get \(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)\), which contains 3 consecutive factors. There would be 4 factors in \(\frac{(n+4)!}{n!}\). Exam tip: In \(\frac{(n+r)!}{n!}\), exactly \(r\) consecutive factors remain.

Correct answer: A. (\frac{12}{5})

The direct answer is option A, \(\frac{12}{5}\). A factorial means multiplying a positive integer by every smaller positive integer: \(9!=9\times8\times7!\), so \(\frac{9!}{7!}=9\times8=72\). Similarly, \(6!=6\times5\times4!\), so \(\frac{6!}{4!}=6\times5=30\). Therefore, \(\frac{x}{y}=\frac{72}{30}=\frac{12}{5}\) after dividing numerator and denominator by 6. Option A is correct because it is the simplified ratio. Option B, \(\frac{5}{12}\), is the reciprocal and would represent \(\frac{y}{x}\), not \(\frac{x}{y}\). Option C, \(\frac{6}{5}\), comes from an incomplete or incorrect cancellation and does not equal \(72\div30\). Option D, \(\frac{5}{6}\), is also a reciprocal-type result and is not the required ratio. Remember: cancel factorials first, then divide x by y in the order given.

Correct answer: A. ((n+1)(n+3))

Factorial notation allows us to cancel common factorial factors by expanding only as far as needed. We use (n+2)!=(n+2)(n+1)n! and (n+1)!=(n+1)n!. This turns both fractions into ordinary algebraic terms. The simplification assumes the factorials are defined, normally for non-negative integer n.

The first fraction becomes (n+2)(n+1)n!/n!=(n+2)(n+1). The second becomes (n+1)n!/n!=n+1. Adding gives (n+1)(n+2)+(n+1). Taking the common factor n+1 gives (n+1)[(n+2)+1]=(n+1)(n+3). Therefore option A is correct. The unsimplified terms must not be added by combining their factorial numerators incorrectly.

Correct answer: B. (66)

(\frac{13!}{11!}=13\times12=156) and (\frac{10!}{8!}=10\times9=90), so the difference is (66). Simplify both factorial ratios separately first.

Correct answer: A. \(n(n-1)!\)

By definition, \(n!=n\times(n-1)!\), so option A is correct. In \((n+1)(n-1)!\), the required factor \(n\) is replaced by \(n+1\). Exam tip: expand a factorial by one term to verify such identities.

Correct answer: B. 64

As \(9!=9\times8!\), we get \(9!-8!=9\times8!-8!=8!(9-1)=8\times8!\). Also, \(8!=8\times7!\). Therefore, \(\frac{8\times8!}{7!}=8\times8=64\). Hence, 64 is the correct answer. The value 72 may result from expanding or cancelling factorial terms incorrectly. Exam tip: first factor out the smaller factorial in such expressions.

Correct answer: B. \((n+4)(n+3)(n+2)\)

\((n+4)!=(n+4)(n+3)(n+2)(n+1)!\). Thus, \((n+1)!\) cancels from the numerator and denominator, leaving \((n+4)(n+3)(n+2)\). Option C uses \((n+1)\) in place of \((n+4)\), so it is incorrect. Exam tip: In factorial fractions, expand the larger factorial only until the denominator factorial appears, then cancel common factors.

Correct answer: C. 30

The governing concept is factorial notation: for a non-negative integer k, k! = k(k-1)(k-2) ... 1, and common factorial factors may be cancelled in a quotient. Substitute n = 4 into the expression: (n+2)! / n! = 6! / 4!. Expand only the needed factors rather than calculating both factorials completely: 6! = 6 × 5 × 4!, so 6! / 4! = 6 × 5 = 30. Hence option C is correct. Option B, 24, is merely 4!, while option D, 36, may result from an incorrect multiplication involving 6^2. Option A, 20, does not follow from the factorial cancellation. The reliable method is to cancel n! after writing (n+2)! as (n+2)(n+1)n!.