Factorial notation

Correct answer: B. 7

Since \((n+1)!=(n+1)\cdot n!\), \(\frac{(n+1)!}{n!}=n+1\). Thus, \(n+1=8\), which gives \(n=7\). Option 8 is a close distractor because it is the value of \(n+1\), not of \(n\). Exam tip: In factorial ratios, first expand the larger factorial in terms of the smaller one.

Correct answer: A. −48

The governing concept is factorial cancellation. For positive integers, 9! = 9 × 8 × 7!, so 9!/7! = 9 × 8 = 72. Also, 5! = 5 × 4 × 3 × 2 × 1 = 120. Substituting these values into the expression gives 72 − 120 = −48. Therefore, option A is correct. The negative sign matters because 120 is larger than 72, and the expression subtracts 120 from 72. Option C is only the value of the quotient and ignores the subtraction. Option D is only 5!, while option B has the wrong sign. Cancelling factorials first is faster and avoids calculating 9! as a very large number.

Correct answer: B. 20

Substituting \(n=3\) gives \(\frac{(n+2)!}{n!}=\frac{5!}{3!}\). Since \(5!=5\times4\times3!\), \(\frac{5!}{3!}=5\times4=20\). Option 60 is the value of \(5!\), but the expression must also be divided by \(3!\). Exam tip: Expand the larger factorial to cancel common \(n!\) terms in a factorial fraction.

Correct answer: B. \(5\)

We know that \(6!=6\times5\times4\times3\times2\times1=720\). Hence, \((n+1)!=6!\), so \(n+1=6\) and \(n=5\). If \(n=6\), then \((n+1)!=7!=5040\), not 720. Exam tip: Memorise small factorial values such as \(5!=120\), \(6!=720\), and \(7!=5040\).

Correct answer: C. 144

Since \(6!=6\times5\times4\times3!\), \(\frac{6!}{3!}=6\times5\times4=120\). Also, \(4!=4\times3\times2\times1=24\). Therefore, \(\frac{6!}{3!}+4!=120+24=144\). Option 120 is only the quotient and does not include \(4!\). Exam tip: In factorial fractions, cancel the common factorial part first.

Correct answer: B. 25

The governing concept is cancellation of common factorial factors. Since 5! = 5 × 4 × 3!, the first quotient is 5!/3! = 5 × 4 = 20. Since 5! = 5 × 4!, the second quotient is 5!/4! = 5. Adding the two independently simplified terms gives 20 + 5 = 25. Hence option B is correct. Option A represents only the first quotient and leaves out the second term. Option C may result from an incorrect addition, and option D has no valid factorial calculation behind it. There is no need to expand 5! into 120 twice; cancelling the lower factorial is valid because factorials are non-zero, and it makes the arithmetic shorter and clearer.

Correct answer: B. 2

Using factorial expansion, \((n+2)!=(n+2)(n+1)!\), so \(\frac{(n+2)!}{(n+1)!}=n+2\). Similarly, \(n!=n(n-1)!\), hence \(\frac{n!}{(n-1)!}=n\). Therefore, the expression is \((n+2)-n=2\). Note that \(n+2\) is only the first ratio, not the difference of the two ratios. Exam tip: In factorial fractions, expand the larger factorial only up to the smaller factorial and cancel common factors.

Correct answer: C. (71)

The direct answer is option C, 71. Interpret the commas in the denominator as multiplication: \(\frac{8!}{5!3!}+\frac{6!}{4!2!}\). First, \(\frac{8!}{5!3!}=\frac{8\times7\times6\times5!}{5!\times3\times2\times1}=\frac{8\times7\times6}{6}=56\). Second, \(\frac{6!}{4!2!}=\frac{6\times5\times4!}{4!\times2}=15\). Adding gives \(56+15=71\). Option A, 56, is only the first term and forgets the second. Option B, 65, results from an arithmetic or factorial error. Option C, 71, is correct. Option D, 84, also does not follow from either the correct simplification or addition. Factorials mean products down to 1, and common factorial factors should be cancelled before multiplying. Memory cue: calculate each fraction separately, then add.

Correct answer: B. 7

Since \(n! = n(n-1)(n-2)!\), \(\frac{n!}{(n-2)!}=n(n-1)\). Hence, \(n(n-1)=42\). The consecutive positive integers \(6\) and \(7\) have product \(42\), so \(n=7\). Option \(6\) is a close distractor, but \(6\times5=30\), not 42. Exam tip: in factorial fractions, first cancel the common factorial term and write the remaining consecutive factors.

Correct answer: C. 102

Use the factorial identity n! = n(n−1)! and cancel the common lower factorial in each quotient. Because 10! = 10 × 9 × 8!, we get 10!/8! = 10 × 9 = 90. Similarly, 4! = 4 × 3 × 2!, so 4!/2! = 4 × 3 = 12. The complete expression is therefore 90 + 12 = 102, making option C correct. Option A is only the first quotient and omits the second one. Option B reflects an incorrect value or addition for the second quotient, while option D is larger than the valid sum. Expanding 10! and 4! completely is unnecessary; cancellation preserves the value and greatly reduces the chance of arithmetic errors.

Correct answer: B. 6

The key idea is to express the larger factorial using the smaller factorial and then factor. Since 7! = 7 × 6!, the numerator becomes 7 × 6! − 6! = (7−1) × 6! = 6 × 6!. Dividing by 6! gives (6 × 6!)/6! = 6, because 6! is non-zero. Thus option B is correct. An equivalent method is to divide each term in the numerator by 6!: 7!/6! − 6!/6! = 7 − 1 = 6. Option C results from forgetting the subtraction of 6!, while option A uses an incorrect subtraction and option D has no valid factorial basis. Factorising before division makes the structure transparent.

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

Expanding the factorial gives \((n+3)!=(n+3)(n+2)(n+1)n!\). On dividing by \(n!\), the factor \(n!\) cancels, leaving \((n+3)(n+2)(n+1)\). Option D has only two factors, whereas three consecutive factors occur after \(n!\) in \((n+3)!\). Exam tip: In factorial quotients, expand the larger factorial up to the smaller one and cancel common factors.

Correct answer: B. 4

Cancelling \(n!\) gives \(\frac{(n+2)!}{n!}=(n+2)(n+1)\). Hence, \((n+2)(n+1)=30\). For \(n=4\), we get \(6\times5=30\), so the correct value is \(4\). Although \(n=-7\) also makes the product 30, ordinary factorial notation requires \(n\) to be a non-negative integer. Exam tip: Expand the larger factorial into consecutive factors first, then cancel the common factorial.

Correct answer: A. (63)

(\frac{9!}{6!,3!}=84) and (\frac{7!}{5!,2!}=21), so the difference is (63). Evaluate both combination-like terms separately.

Correct answer: C. \(\frac{21}{4}\)

Here, \(5!=120\), \(3!=6\), and \(4!=24\). Therefore, \(\frac{5!+3!}{4!}=\frac{120+6}{24}=\frac{126}{24}=\frac{21}{4}\). Hence, \(\frac{21}{4}\) is the correct answer. Option 5 is only the value of \(\frac{5!}{4!}\); it ignores the \(3!\) term. Exam tip: evaluate each factorial first, then perform the addition and division.

Correct answer: B. (1470)

(\frac{8!-7!}{4!}=\frac{7!(8-1)}{4!}=7\times7\times6\times5=1470). With a common denominator, combine the numerator.

Correct answer: B. 55

Since \(11!=11\times10\times9!\) and \(2!=2\), \(\frac{11!}{9!,2!}=\frac{11\times10\times9!}{9!\times2}=\frac{110}{2}=55\). The value 66 is \(\binom{12}{2}\), so it does not apply here. Exam tip: In factorial fractions, first cancel the common factorial terms and then multiply the remaining factors.

Correct answer: B. \(12\times11\times10\)

Write \(12!=12\times11\times10\times9!\). Then \(\frac{12!}{9!}=\frac{12\times11\times10\times9!}{9!}=12\times11\times10\). Hence, option B is correct. Option D wrongly includes 9 instead of 12. Exam tip: for \(\frac{n!}{r!}\), where \(n>r\), the factors from \(n\) down to \(r+1\) remain after cancellation.

Correct answer: B. \(n\)

Using \((n+1)!=(n+1)n!\), the numerator becomes \((n+1)n!-n!=n![(n+1)-1]=nn!\). Hence, \(\frac{nn!}{n!}=n\), so option B is correct. The distractor \(n+1\) would result if the subtraction of \(n!\) were ignored. Exam tip: In factorial expressions, first rewrite the larger factorial as a multiple of the smaller factorial.

Correct answer: C. \(7\)

Since \(n!=n\times(n-1)!\), we get \(\frac{n!}{(n-1)!}=n\) (for \(n\geq1\)). Thus, the given equation becomes \(n+2=9\), which gives \(n=7\). Option \(8\) is a close distractor, but it makes the left-hand side \(8+2=10\), not 9. Exam tip: In a fraction involving consecutive factorials, expand the larger factorial in terms of the smaller one before simplifying.

Correct answer: B. (25)

The direct answer is option B, 25. Factorials can be cancelled in combinations of this form. First, \(\frac{6!}{2!4!}=\frac{6\times5\times4!}{2!\times4!}=\frac{6\times5}{2}=15\), since \(4!\) cancels and \(2!=2\). Next, \(\frac{5!}{2!3!}=\frac{5\times4\times3!}{2!\times3!}=\frac{5\times4}{2}=10\). Adding the two separately calculated values gives \(15+10=25\). Option A, 20, is wrong because it does not give the correct sum. Option B, 25, is correct. Option C, 30, may result from an arithmetic or factorial-cancellation error. Option D, 35, is also inconsistent with the two correct terms. Do not combine the fractions incorrectly; simplify each fraction first and add only at the end. Memory cue: cancel common factorial parts before ordinary multiplication and addition.

Correct answer: B. 70

Here, \(\frac{7!}{3!\,4!}=\binom{7}{3}=35\). Also, \(2!=2\), so the required value is \(35\times2=70\). Option 35 is only the value of the fraction; multiplication by \(2!\) is still required. Exam tip: In factorial expressions, cancel common factorial terms before calculating.

Correct answer: B. 81

Since \(10!=10\times9!\), the numerator becomes \(10!-9!=10\times9!-9!=9!(10-1)=9\times9!\). Hence, \(\frac{9\times9!}{8!}=9\times9=81\). Therefore, 81 is the correct answer. Option 90 can result from calculating \(10!/8!\) while incorrectly ignoring \(-9!\). Exam tip: In factorial expressions, rewrite the larger factorial in terms of the smaller one and factor out the common term first.

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

Using \((n+2)!=(n+2)(n+1)!\), the numerator becomes \((n+1)![(n+2)-1]=(n+1)(n+1)!\). Now, \((n+1)!=(n+1)n!\), so the numerator is \((n+1)^2n!\). Dividing by \(n!\) gives \((n+1)^2\). The expression \(n(n+1)\) would result from the incorrect step \((n+1)!=n\cdot n!\). Exam tip: rewrite factorial terms in terms of a common smaller factorial, such as \(n!\), before simplifying.

Correct answer: B. 5

Write \((n+3)!=(n+3)(n+2)(n+1)!\). Cancelling \((n+1)!\) gives \((n+3)(n+2)=56\). The consecutive factors of 56 are 7 and 8, so \(n+2=7\) and hence \(n=5\). For the close distractor 4, \((4+3)(4+2)=7\times6=42\), not 56. Exam tip: In a factorial fraction, expand the larger factorial only as far as the smaller factorial, then cancel common factors.