Factorial notation

Since \(10!=10\cdot9\cdot8!\), \(\frac{10!}{8!\cdot2!}=\frac{10\cdot9\cdot8!}{8!\cdot2}=\frac{90}{2}=45\). Hence, 45 is correct. The close distractor 90 results from forgetting to divide by \(2!\). Exam tip: expand the larger factorial only up to the smaller factorial, then cancel common factors.

Since \((n+2)!=(n+2)(n+1)n!\), we get \(\frac{(n+2)!}{n!}=\frac{(n+2)(n+1)n!}{n!}=(n+2)(n+1)\). Therefore, option C is correct. \(n+2\) is only one factor and misses the factor \((n+1)\). Exam tip: In a factorial ratio, expand the larger factorial up to the smaller one and then cancel common factors.

(\frac{(n+1)!}{(n-1)!}=n(n+1)), so (n(n+1)=30) gives (n=5). Reduce factorials first in such questions.

(\frac{11!}{9!}=110) and (\frac{8!}{6!}=56), so the difference is (110-56=54). Do the subtraction carefully after simplification.

\(6!=720\) and \(3!=6\). Therefore, \(\frac{6!}{3!\cdot3!}=\frac{720}{6\cdot6}=\frac{720}{36}=20\). The value 15 can result from using incorrect terms in a combination formula. Exam tip: evaluate the factorials first, then divide by the product in the denominator.

The direct answer is option C, 66. Expand only the part of 12! that is not contained in 10!: \(12!=12\times11 imes10!\). Thus, \(\frac{12!}{10! imes2}=\frac{12 imes11 imes10!}{10! imes2}=\frac{12 imes11}{2}=6 imes11=66\). Option A, 54, is wrong because it does not result from the correct remaining factors. Option B, 60, may come from using 10 instead of 11 or making an arithmetic error. Option C, 66, is correct. Option D, 72, is the product 12 × 6 and does not represent division by 2 after the correct cancellation. The key is that the denominator contains both 10! and 2; cancel 10! first, not the numbers 10 or 12 separately. Memory cue: write factorials as products before cancelling.

(n!) means the product of all natural numbers from (1) to (n). A clear definition makes simplification easier.

Factorials are products of consecutive positive integers. When one factorial is divided by a smaller factorial, expand the larger factorial only until the denominator’s factorial appears. This exposes the common factor and makes cancellation straightforward, avoiding unnecessary calculation of the very large values of 13! and 11!.

We write \(13!=13\times12\times11!\). Hence \(\frac{13!}{11!}=\frac{13\times12\times11!}{11!}=13\times12\). The common factor \(11!\) cancels, leaving the product shown in option B. Option C uses 11 instead of the factor immediately above the denominator, while the other choices do not represent the correct factorial expansion.

The numerator is (24+6=30) and denominator is (6+2=8), so the value is (\frac{15}{4}). If options do not match, recalculation is necessary first.

The useful factorial rule is \(m!=m(m-1)!\). Apply it separately to each fraction rather than combining the two fractions at the beginning. In the first fraction, \((n+2)!=(n+2)(n+1)!\), so division by \((n+1)!\) leaves \(n+2\). In the second, \((n+1)!=(n+1)n!\), so division by \(n!\) leaves \(n+1\).

The original expression therefore becomes \((n+2)+(n+1)\). Combining like terms gives \(n+n+2+1=2n+3\). Thus option C is correct. The answer is not \(2n+2\), because the constant terms 2 and 1 add to 3. The simplification assumes the factorials are defined, as usual in this context, for the relevant nonnegative integer values of \(n\).

Write \(7!=7\cdot6\cdot5\cdot4!\). Therefore, \(\frac{7!}{4!}=\frac{7\cdot6\cdot5\cdot4!}{4!}=7\cdot6\cdot5\). Option B incorrectly includes an extra \(4\), since \(4!\) has already been cancelled from the denominator. Exam tip: for \(\frac{n!}{r!}\), where \(n>r\), write the factors from \(n\) down to \(r+1\).

Cancelling \(n!\) gives \(\frac{(n+2)!}{n!}=(n+2)(n+1)\). Hence, \((n+2)(n+1)=56\). The consecutive factors of 56 are \(7\) and \(8\), so \(n+1=7\) and \(n=6\). Although \(n=-9\) is an algebraic root, factorials are not defined for negative integers. Exam tip: In such factorial fractions, expand up to the smaller factorial and cancel first.

The direct answer is option C, 84. Since \(9!=9 imes8 imes7 imes6!\), we have \(\frac{9!}{6!}=9 imes8 imes7\). The expression becomes \(\frac{9 imes8 imes7}{3!}\). Because \(3!=3 imes2 imes1=6\), the value is \(\frac{9 imes8 imes7}{6}=84\). Option A, 56, is wrong because the numerator has not been divided correctly by 6. Option B, 72, is also an arithmetic error. Option C, 84, is correct. Option D, 96, does not follow from the factorial cancellation and is too large. A useful check is that 9×8×7 = 504 and 504 ÷ 6 = 84. Do not treat 3! as 3; it equals 6. Exam cue: cancel the larger common factorial first, then evaluate the small remaining factorial.

Since \((n+1)!=(n+1)\,n\,(n-1)!\), dividing by \((n-1)!\) gives \(\frac{(n+1)!}{(n-1)!}=n(n+1)\). In option D, the factor \(n\) is missing. Exam tip: In a factorial fraction, expand the larger factorial up to the smaller factorial and then cancel common factors.

Since (5!=5\cdot4!), (\frac{4!\cdot5}{5!}=1). Write a factorial using the nearest smaller factorial.

\(3!=3\times2\times1=6\) and \(4!=4\times3\times2\times1=24\). Therefore, \(3!\cdot4!=6\times24=144\). Note that 120 is the value of \(5!\), so it is not the required product. Exam tip: evaluate each factorial first, then multiply the results.

Since \(8! = 8\cdot7\cdot6\cdot5!\) and \(3!=6\), \(\frac{8!}{5!\cdot3!}=\frac{8\cdot7\cdot6\cdot5!}{5!\cdot6}=8\cdot7=56\). Therefore, 56 is correct. The value 48 can result from not dividing correctly by \(3!\). Exam tip: In factorial fractions, cancel the common factorial first and then multiply the remaining factors.

Since 120 = 5!, we get n = 5. Therefore, (n-2)! = (5-2)! = 3! = 3 × 2 × 1 = 6. The value 24 is 4!, which would arise if (n-1)! were asked. Exam tip: First identify n from the given factorial, then substitute it into the required factorial expression.

Since \(7!=7\times6\times5\times4!\), \(\frac{7!}{3!\cdot4!}=\frac{7\times6\times5}{3!}=\frac{7\times6\times5}{6}=35\). Therefore, 35 is the correct option. Getting 42 is a common error caused by not dividing correctly by \(3!=6\). Exam tip: In factorial fractions, cancel common factorial terms first and then multiply the remaining factors.

Here, \(5!=5\times4\times3\times2\times1=120\) and \(2!=2\). Therefore, \(\frac{5!}{2!}=\frac{120}{2}=60\). Also, \(1!=1\), so \(\frac{5!}{2!}+1!=60+1=61\). Option 60 results from forgetting to add \(1!\). Exam tip: Evaluate each factorial first, then perform the remaining operations.

The governing concept is the consecutive-factorial identity k! = k(k−1)!. Assuming the factorials are defined, simplify each quotient separately. First, (n+2)!/(n+1)! = [(n+2)(n+1)!]/(n+1)! = n+2. Second, n!/(n−1)! = [n(n−1)!]/(n−1)! = n. Subtracting gives (n+2) − n = 2, so option B is correct. Option C is only the value of the second quotient, and option D is only the value of the first quotient. Option A incorrectly removes the added 2 instead of retaining it. The result is constant because the variable terms cancel, which is the important algebraic feature of the expression.

Factorial is defined for non-negative integers, so (n-3\ge0). Among the given options, (n=3) is correct.

Factorials should be evaluated before combining the terms. The factorial \(n!\) is the product from \(n\) down to 1, so \(2!=2\), \(3!=6\), and \(4!=24\). The dot between the first two factorials means multiplication, and multiplication is performed before the addition. Thus the expression has value 36, which is option B.

Calculate the product first: \(2!\cdot3!=2\cdot6=12\). Then calculate the remaining factorial, \(4!=4\cdot3\cdot2\cdot1=24\). Add the two results: \(12+24=36\). The answer is not 30, 42, or 48 because those values result from an arithmetic or order-of-operations mistake. Therefore option B follows.

Using factorial notation, \((n+3)!=(n+3)(n+2)(n+1)n!\). Dividing by \(n!\) cancels \(n!\), leaving \((n+3)(n+2)(n+1)\). Option A misses the factor \((n+1)\). Exam tip: In a factorial quotient, expand the larger factorial only up to the smaller factorial and then cancel common factors.

(\frac{10!}{7!\cdot3!}=\frac{10\cdot9\cdot8}{6}=120). Reduce the large factorial to three factors.