Factorial notation

Correct answer: D. 49

Writing \(8!=8\times7!\), we get \(8!-7!=8\times7!-7!=7!(8-1)=7\times7!\). Since \(7!=7\times6!\), \(\frac{7\times7!}{6!}=\frac{7\times7\times6!}{6!}=49\). Therefore, 49 is correct. The value 56 may result from failing to express \(7!\) correctly as \(7\times6!\). Exam tip: in factorial expressions, first factor out the smaller factorial common to the terms.

Correct answer: B. \(8\)

Using the factorial definition, \(n!=n(n-1)(n-2)!\). Therefore, \(\frac{n!}{(n-2)!}=n(n-1)=56\). Since \(8\times7=56\), \(n=8\). The close distractor \(7\) gives \(7\times6=42\), not 56. In factorial notation, \(n\) is taken as a positive integer, so a negative algebraic root is not valid. Exam tip: Expand the larger factorial only up to the smaller factorial before cancelling.

Correct answer: D. 35

Since \(6!=6\times5\times4!\) and \(5!=5\times4!\), \(\frac{6!+5!}{4!}=\frac{(6\times5\times4!)+(5\times4!)}{4!}=30+5=35\). Hence, 35 is correct. The value 30 comes from considering only the \(6!\) term; the \(5!\) term must also be included. Exam tip: Rewrite larger factorials as multiples of a common smaller factorial before simplifying.

Correct answer: A. (462)

The direct answer is option A, 462. Simplify the first quotient by cancelling 6!: \(\frac{9!}{6!}=\frac{9\times8\times7\times6!}{6!}=9\times8\times7=504\). For the second quotient, \(\frac{7!}{5!}=\frac{7\times6 imes5!}{5!}=7\times6=42\). The required subtraction is therefore \(504-42=462\). Option A, 462, is correct. Option B, 480, would come from an incorrect subtraction or from failing to reduce one quotient properly. Option C, 504, is only the first quotient and ignores the subtraction of 42. Option D, 546, does not follow from the required difference and may reflect adding rather than subtracting. The safe method is to cancel the smaller factorial first, calculate each quotient, and then apply the indicated minus sign. Exam cue: \(\frac{n!}{r!}\) leaves the factors from n down to r+1.

Correct answer: B. 2(n+2)^2

The governing concept is factorial expansion and algebraic factorisation. Expand each quotient only until the denominator cancels: (n+3)!/(n+1)! = (n+3)(n+2), because (n+3)! = (n+3)(n+2)(n+1)!; similarly, (n+2)!/n! = (n+2)(n+1). Add the results: (n+3)(n+2) + (n+2)(n+1). Taking the common factor (n+2) gives (n+2)[(n+3)+(n+1)] = (n+2)(2n+4) = 2(n+2)(n+2) = 2(n+2)^2. Therefore option B is correct. Option A misses the factor 2, option C is only the bracket factor and omits multiplication by (n+2), and option D does not represent the correct expansion.

Correct answer: C. \(5\)

Since \((n+2)!=(n+2)(n+1)n!\), we get \(\frac{(n+2)!}{n!}=(n+2)(n+1)\). Hence, \((n+2)(n+1)=42\). The consecutive positive integers whose product is \(42\) are \(6\) and \(7\), so \(n+1=6\) and \(n=5\). If \(n=4\), the product would be \(5\times6=30\), so it is not correct. Exam tip: Before cancelling a factorial in a fraction, expand the larger factorial up to the common factorial term.

Correct answer: D. 94

Here, the comma in each denominator indicates multiplication. Thus, \(\frac{12!}{10!\,2!}=\frac{12\times11}{2}=66\) and \(\frac{8!}{6!\,2!}=\frac{8\times7}{2}=28\). Therefore, the sum is \(66+28=94\). Option 90 can result from an error in multiplication or addition. Exam tip: Cancel common factorial terms first to simplify quickly.

Correct answer: A. \(n^2\)

Using \((n+1)!=(n+1)n!\), the numerator becomes \((n+1)n!-n!=n![(n+1)-1]=n\cdot n!\). Since \(n!=n(n-1)!\), \(\frac{n\cdot n!}{(n-1)!}=n\cdot n=n^2\). Hence, the correct answer is \(n^2\). The expression \(n(n+1)\) would result from incorrectly ignoring the subtraction of \(n!\). Exam tip: In factorial expressions, first rewrite the larger factorial in terms of the smaller one.

Correct answer: C. 210

Here, the denominator is intended to mean the product \(4!\,3!\). Thus, \(\frac{7!}{4!\,3!}\times3!=\frac{7!}{4!}=7\times6\times5=210\). Therefore, \(210\) is the correct option. An answer such as \(140\) may result from not cancelling \(3!\) correctly. Exam tip: Cancel common factors in the numerator and denominator before expanding factorials.

Correct answer: C. 76

\(\frac{8!}{6!}=\frac{8\times7\times6!}{6!}=8\times7=56\), and \(\frac{5!}{3!}=\frac{5\times4\times3!}{3!}=5\times4=20\). Hence, \(x+y=56+20=76\). 72 may seem close, but it can result from simplifying \(8!\) incorrectly against \(6!\). Exam tip: In a factorial fraction, cancel the common factorial term first and multiply only the remaining factors.

Correct answer: A. \((n+5)(n+4)\)

Expand the larger factorial up to the smaller one: \((n+5)!=(n+5)(n+4)(n+3)!\). The common factor \((n+3)!\) cancels from the numerator and denominator, giving \(\frac{(n+5)!}{(n+3)!}=(n+5)(n+4)\). Option B incorrectly uses \(n+3\) in place of \(n+4\). Exam tip: In factorial division, expand the larger factorial only until the smaller factorial appears, then cancel common factors.

Correct answer: C. 100

Since \(11!=11\times10!\), we get \(11!-10!=11\times10!-10!=10!(11-1)=10\times10!\). Also, \(10!=10\times9!\), so \(\frac{10\times10!}{9!}=10\times10=100\). Hence, 100 is the correct answer. The value 110 can result from not simplifying the subtraction involving \(11!\) and \(10!\) correctly. Exam tip: When factorial terms are subtracted, first factor out the smaller factorial.

Correct answer: B. 12

\(\frac{9!}{7!}=\frac{9\times8\times7!}{7!}=9\times8=72\), and \(\frac{6!}{5!}=6\). Therefore, the given value is \(72\div6=12\). A value such as \(18\) can result from cancelling factorial factors incorrectly. Exam tip: In factorial fractions, cancel the common factorial part first and then multiply the remaining factors.

Correct answer: C. 8

Since \((n+1)!=(n+1)n(n-1)!\), the factor \((n-1)!\) cancels, so the given expression becomes \(n(n+1)\). Thus, \(n(n+1)=72\). The consecutive integers whose product is 72 are 8 and 9; hence, \(n=8\). The closest option, 7, gives \(7\times8=56\), not 72. Exam tip: expand the larger factorial up to the smaller factorial and then cancel common factors.

Correct answer: C. 15

Taking the commas in the given expression to indicate multiplication, \(\frac{5!\times3!}{2!\times4!}=\frac{5!}{4!}\times\frac{3!}{2!}=5\times3=15\). Therefore, the correct answer is 15. Option 20 can result from incorrect cancellation of factorial terms. Exam tip: simplify matching factorials quickly using \(\frac{n!}{(n-1)!}=n\).

Correct answer: D. 342

\(\frac{8!}{5!}=8\times7\times6=336\), since \(5!\) cancels out. Similarly, \(\frac{6!}{4!}=6\times5=30\), and \(4!=24\). Therefore, the value is \(336+30-24=342\). Choosing \(336\) would ignore the effect of the second term and \(4!\). Exam tip: In factorial fractions, cancel the common factorial first and then multiply the remaining factors.

Correct answer: B. 3

Since \((n+5)!=(n+5)(n+4)(n+3)(n+2)!\), cancelling \((n+2)!\) leaves \((n+5)(n+4)(n+3)\). Hence, 3 consecutive factors remain. Four factors would remain if the denominator were \((n+1)!\). Exam tip: in \(\frac{(n+r)!}{n!}\), usually \(r\) factors remain after cancellation.

Correct answer: B. \(7\)

Cancelling \((n+1)!\) from \((n+3)!\) gives \(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\). Thus, \((n+3)(n+2)=90\). The consecutive factors are \(9\) and \(10\), so \(n+2=9\) and \(n=7\). If \(n=8\), the product would be \(10\times11=110\), not \(90\). Exam tip: In factorial fractions, first cancel the common ending factorial terms.

Correct answer: C. 63

\(\frac{13!}{11!\,2!}=\frac{13\times12}{2}=78\), while \(\frac{6!}{4!\,2!}=\frac{6\times5}{2}=15\). Therefore, the required value is \(78-15=63\). The option 60 can result from incorrect cancellation of factorial terms. Exam tip: use \(\frac{n!}{(n-2)!\,2!}=\frac{n(n-1)}{2}\) to evaluate such terms quickly.

Correct answer: B. \(n^2+3n+1\)

Since \((n+2)!=(n+2)(n+1)n!\), \(\frac{(n+2)!-n!}{n!}=(n+2)(n+1)-1\). Thus, it becomes \(n^2+3n+2-1=n^2+3n+1\). Option \(n^2+3n\) misses the constant term remaining after subtracting 1. Exam tip: In factorial fractions, first expand the larger factorial in terms of the smaller one.

Correct answer: C. 98

\(\frac{10!}{8!}=\frac{10\times9\times8!}{8!}=90\) and \(\frac{7!}{6!}=\frac{7\times6!}{6!}=7\). Also, by definition, \(0!=1\). Therefore, the value is \(90+7+1=98\). One may get \(97\) by incorrectly taking \(0!\) as 0. Exam tip: In factorial fractions, expand only until the common factorial terms cancel.

Correct answer: C. \(n(n+1)\)

Using factorial notation, \(n!=n(n-1)!\), so \(\frac{n!}{(n-1)!}=n\). Likewise, \((n+1)!=(n+1)n!\), so \(\frac{(n+1)!}{n!}=n+1\). Hence the full expression is \(n\times(n+1)=n(n+1)\). \(n^2\) would result only if both factors were \(n\). Exam tip: In factorial ratios, expand the larger factorial just enough to cancel the smaller one.

Correct answer: B. 7

Since \(n!=n(n-1)(n-2)(n-3)!\), we get \(\frac{n!}{(n-3)!}=n(n-1)(n-2)\). Thus, \(n(n-1)(n-2)=210\). As \(7\times6\times5=210\), \(n=7\). For example, if \(n=8\), the product is \(8\times7\times6=336\), not 210. Exam tip: In factorial quotients, cancel the denominator factorial and write the remaining consecutive factors.

Correct answer: B. 80

Using \(9!=9\times 8!\), the numerator becomes \(9\times 8!+8!=8!(9+1)=10\times 8!\). Therefore, \(\frac{10\times 8!}{7!}=10\times 8=80\), since \(8!=8\times 7!\). Hence, 80 is correct. The value 72 results from dividing only \(9!\) by \(7!\) and incorrectly omitting the \(8!\) term. Exam tip: factor out the common factorial before simplifying.

Correct answer: C. 41

\(\frac{6!}{3!\,3!}=\frac{720}{6\times6}=20\), and \(\frac{7!}{5!\,2!}=\frac{5040}{120\times2}=21\). Therefore, the total is \(20+21=41\). An option such as \(39\) may result from an error in simplification or addition. Exam tip: Cancel factorial terms first and write the remaining products carefully.