Factorial notation

The governing concept is factoring and cancelling factorials. Write 6! as 6 × 5!. Then the numerator becomes 6 × 5! − 5! = (6−1) × 5! = 5 × 5!. Since 5! = 5 × 4!, the entire expression is [5 × (5 × 4!)]/4! = 25. Thus option C is correct. Direct evaluation also confirms this: 6! − 5! = 720 − 120 = 600, and 600/24 = 25, but the cancellation method is shorter. Option B may come from confusing the numerator with 5!, option D from retaining 6 instead of 5 after subtraction, and option A has no valid factorial basis.

Since \((n+2)!=(n+2)(n+1)!\), the numerator becomes \((n+2)(n+1)!+(n+1)!\). Taking \((n+1)!\) common gives \([(n+2)+1](n+1)!=(n+3)(n+1)!\). Dividing by \((n+1)!\) gives \(n+3\). Option \(n+2\) misses the contribution of the second term, \((n+1)!\). Exam tip: in factorial expressions, factor out the smaller factorial first.

(\frac{12!}{9!}=12\cdot11\cdot10), so three factors remain. Cancel the numerator up to the denominator factorial.

The intended meaning of the given expression is \(r! = 24\). Since \(4! = 4 \times 3 \times 2 \times 1 = 24\), we get \(r=4\). Therefore, \((r+1)! = 5! = 5 \times 4! = 5 \times 24 = 120\). The value \(720\) is \(6!\), so it is not correct. Exam tip: Use \((n+1)!=(n+1)\,n!\) to solve such questions quickly.

Using the factorial identity, \((n+1)!=(n+1)\times n!\). Therefore, the numerator and denominator are equal, so \(\frac{(n+1)!}{(n+1)n!}=1\). \(n+1\) is not the answer because it is only one factor in the expanded numerator; \(n!\) is also present in the denominator. Exam tip: In factorial fractions, first expand the larger factorial in terms of the smaller one and then cancel common factors.

The denominator is (n-1)!. Rewrite the numerator using n!=n(n-1)!: n!+(n-1)!=n(n-1)!+(n-1)!=(n+1)(n-1)!. Dividing by (n-1)! gives n+1, assuming the factorial is defined and nonzero. The equation therefore becomes n+1=9, so n=8.

Hence option C is correct. The important step is factoring the common factorial before cancelling it. A direct calculation of large factorials is unnecessary and can cause errors. The values 6, 7, and 9 would make the simplified expression equal to 7, 8, and 10 respectively, not 9. Thus the algebra supports only 8 among the listed choices.

(\frac{14!}{12!\cdot2!}=\frac{14\cdot13}{2}=91). Expand the larger factorial only up to the smaller factorial.

(\frac{15!}{13!}=15\cdot14=210) and (5!=120), so the difference is (90). Simplify the factorial ratio first.

Cancel \(8!\) from \(11!\): \(\frac{11!}{8!\cdot3!}=\frac{11\cdot10\cdot9}{3\cdot2\cdot1}=\frac{990}{6}=165\). Hence, 165 is the correct option. An answer such as 180 can result from not dividing correctly by \(3!\). Exam tip: In factorial fractions, expand the larger factorial only up to the smaller factorial, then cancel common factors.

The numerator is (7!+6!=8\cdot6!) and the denominator is (6!+5!=7\cdot5!), so careful direct calculation is needed. Substitute values to avoid mistakes.

Since \(18!=18\cdot17\cdot16!\), \(\frac{18!}{16!\cdot2}=\frac{18\cdot17\cdot16!}{16!\cdot2}=\frac{18\cdot17}{2}=9\cdot17=153\). Therefore, the correct answer is 153. The value 144 would result from using \(18\times16/2\), which incorrectly omits the factor 17. Exam tip: In factorial fractions, expand the larger factorial up to the smaller factorial and cancel common factors.

(\frac{(n+4)!}{(n+1)!}=(n+4)(n+3)(n+2)), so three factors remain. Reduce the numerator up to the denominator factorial.

\(3!=6\), \(2!=2\), and \(1!=1\). Therefore, the numerator is \(6+2+1=9\). Also, \(0!=1\), so \(\frac{9}{1}=9\). Option 8 may result from incorrectly taking \(1!\) as 0. Exam tip: Always remember that both \(0!\) and \(1!\) are equal to 1.

Use \((n+3)!=(n+3)(n+2)!\). The numerator becomes \((n+3)(n+2)!-(n+2)!=[(n+3)-1](n+2)!=(n+2)(n+2)!\). Dividing by \((n+2)!\) gives \(n+2\). Choosing \(n+3\) would ignore the subtraction of the second factorial term. Exam tip: when factorial terms differ by one factor, factor out the smaller factorial.

Since \(5040=7!\), we get \(a=7\). Therefore, \((a-4)!=(7-4)!=3!=6\). The option 24, which is \(4!\), may result from incorrectly taking \(a-4\) as 4. Exam tip: first determine the variable from the given factorial, then substitute it into the required expression.

(\frac{16!}{14!}=16\cdot15=240) and (\frac{5!}{4!}=5), so the sum is (245). Simplify the ratios separately.

Since \((n+1)!=(n+1)\,n\,(n-1)(n-2)!\), dividing by \((n-2)!\) gives \(\frac{(n+1)!}{(n-2)!}=(n+1)n(n-1)\). Option A misses the factor \((n-1)\). Exam tip: In factorial division, expand the numerator only up to the factorial in the denominator and then cancel common factors.

(\frac{(m+2)!}{m!}=(m+2)(m+1)), so (10\cdot9=90) gives (m=8). Identify the product of consecutive numbers.

(\frac{8!}{4!\cdot2!\cdot2!}=\frac{8\cdot7\cdot6\cdot5}{4}=420). Cancel (4!) first and calculate with smaller factors.

(72=2^3\cdot3^2), and (6!) contains these factors. In such questions, use prime factorization of the number.

The governing idea is factorial cancellation. A factorial can be written as a product of the required final factors followed by the smaller factorial: 11!=11×10×9!. Also, 2!=2×1=2. Therefore, 11!/(9!×2!)=(11×10×9!)/(9!×2)=(11×10)/2=110/2=55. Thus option B is correct. The expression also represents the combination 11C2, but no combination formula is needed; direct cancellation is the clearest method. Option A results from forgetting to divide by 2!, while option C does not follow from valid cancellation. Option D can arise from an incorrect reduction of 11×10. The factor 9! cancels only with the matching factor in the numerator, not with the entire denominator.

Use the factorial identity 15!=15×14×13!. Substituting this form into the expression gives 15!/(13!×2)=(15×14×13!)/(13!×2). The common factor 13! cancels, leaving (15×14)/2=210/2=105. Therefore option A is correct. The expression is also equal to the combination 15C2, although recognizing that interpretation is optional. Option B is the product 15×14 before division by 2, so it ignores an essential denominator factor. Option C is unrelated to the required cancellation, and option D may result from an incorrect arithmetic operation. The denominator 2 remains after 13! is cancelled and must not be discarded.

The governing concept is factoring a common factorial from a difference. Express 13! as 13 × 12!, so 13! − 12! = 13 × 12! − 12! = 12 × 12!. Now use 12! = 12 × 11!, which gives 12 × 12! / 11! = 12 × (12 × 11!) / 11! = 12 × 12 = 144. Hence option A is correct. Equivalently, dividing each factorial by 11! first gives 13×12 − 12 = 156−12 = 144. Option B subtracts incorrectly, option C corresponds to the first expanded product without the subtraction, and option D is not obtained from the factorial identities.

The governing method is factorial cancellation followed by subtraction. For the first term, 14!/(12!·2!)=(14×13)/(2×1)=182/2=91. For the second term, 13!/(11!·2!)=(13×12)/(2×1)=156/2=78. The required expression is therefore 91−78=13, so option A is correct. Another efficient check is to recognize the two terms as 14C2 and 13C2, whose difference is 91−78. Option B would come from adding equal-looking differences rather than evaluating the stated subtraction. Options C and D result from incorrect combinations of the intermediate products. The 2! denominator must be retained in both terms before subtraction.

Use the factorial identity m!/(m−2)!=m(m−1). With m=2n, the given equation becomes (2n)(2n−1)=132. We need two consecutive positive factors whose product is 132. Since 12×11=132 and the even factor must be 2n, we get 2n=12, hence n=6. Direct verification gives 12!/10!=12×11=132. Therefore option B is correct. If n=5, the product is 10×9=90; if n=7, it is 14×13=182; and if n=8, it is 16×15=240. Thus none of the other options satisfies the original equation. The factorial cancellation is valid when the factorial arguments are defined.