Factorial notation

Use the factorial expansion rule: m! = m(m − 1)(m − 2)!, with the expansion continued until the denominator factorial appears. Here, (n + 2)! = (n + 2)(n + 1)n(n − 1)!. Dividing by (n − 1)! cancels that common factorial and leaves n(n + 1)(n + 2). Thus option A is correct. Option B omits the factor n, option C stops at the wrong lower factor, and option D omits the factor n + 1. The simplification assumes the factorials are defined, so n is an integer for which n − 1 is a nonnegative integer; within that domain, the cancellation is exact.

Answer: 8, so option B is correct. Recall that a factorial is a product of consecutive positive integers: (n+1)! = (n+1)n(n−1)(n−2)!. Cancelling (n−2)! from numerator and denominator gives (n+1)n(n−1)=504. We need three consecutive integers whose product is 504. The numbers 7, 8 and 9 give 7×8×9=56×9=504, and n is the middle number, so n=8. Checking directly, 9!/6!=9×8×7=504. Option A, n=7, gives 8×7×6=336; option C, n=9, gives 10×9×8=720; and option D, n=6, gives 7×6×5=210. Memory cue: when factorials are divided, expand only the larger factorial down to the smaller one; do not calculate the huge factorials separately.

Apply factorial cancellation to each fraction. Since (n+2)!=(n+2)(n+1)n!, the first fraction is (n+2)(n+1). Since (n+1)!=(n+1)n(n−1)!, the second fraction is (n+1)n. Thus the difference is (n+2)(n+1)−n(n+1). Factor out the common factor n+1: (n+1)[(n+2)−n]=(n+1)·2=2(n+1). Hence option A is correct. Option C represents only the second product, option B omits the factor 2, and option D comes from mishandling the subtraction or expanding incorrectly. The factorial expressions require n≥1 for the denominators to be defined.

Cancel the common factorial portion by expanding the numerator to the denominator: (2n+1)!/(2n−1)! = (2n+1)(2n). Therefore, (2n+1)(2n) = 210. The two consecutive factors 15 and 14 multiply to 210. Because 2n is even, it must be 14, while 2n+1 is 15. Thus 2n = 14 and n = 7. Verification gives 15!/13! = 15 × 14 = 210, so option B is correct. The alternatives do not satisfy the equation: n = 6 gives 13 × 12 = 156, n = 8 gives 17 × 16 = 272, and n = 5 gives 11 × 10 = 110. The essential idea is recognizing the consecutive factors after cancellation.

Answer: 49, so option D is correct. Use the factorial rule 8!=8×7! and 7!=7×6!. First factor the numerator: 8!−7!=8×7!−7!=7!(8−1)=7×7!. Therefore (8!−7!)/6!=7×(7!/6!). Since 7!=7×6!, the remaining ratio 7!/6! equals 7. Thus the complete value is 7×7=49. Option A, 35, does not result from the correct cancellation; option B, 42, may come from using only one factor of 7; and option C, 56, incorrectly treats the subtraction or multiplication. Directly calculating 8! and 7! is unnecessary and makes errors more likely. The safe method is to extract a common factorial before dividing.

Expand the numerator only as far as necessary to expose the denominator factors. We have (n + 2)! = (n + 2)(n + 1)n(n − 1)(n − 2)!. Substituting this into the fraction gives [(n + 2)(n + 1)n(n − 1)(n − 2)!]/[(n − 2)! · n(n − 1)]. The common factors (n − 2)!, n, and n − 1 cancel, leaving (n + 2)(n + 1). Therefore option A is correct. Option B is a cancelled denominator part, option C omits the factor n + 2, and option D incorrectly retains n − 1 instead of n + 1. The cancellation requires nonzero denominator factors.

Simplify each factorial quotient separately by cancelling the common factorial in its denominator. For the first term, 16!/(14!·2!) = (16·15·14!)/(14!·2) = (16·15)/2 = 120. For the second term, 15!/(13!·2!) = (15·14·13!)/(13!·2) = (15·14)/2 = 105. Adding the two values gives 120+105 = 225. Therefore option C is correct. The factor 2! equals 2 and must be retained in both denominators; omitting it would double the terms. Option A is only the first term, option B is an incorrect near-total, and option D results from an arithmetic or cancellation mistake.

Use factorial cancellation rather than expanding 13! and 12! completely. Since 13! = 13·12·11·10!, we have 13!/10! = 13·12·11 = 1716. Similarly, 12! = 12·11·10·9!, so 12!/9! = 12·11·10 = 1320. Adding the two simplified values gives 1716+1320 = 3036. Therefore option A is correct. The governing idea is k! = k(k−1)…1, which means the common lower factorial cancels exactly. Option B can result from an addition error, while C and D are not obtained from either correct quotient or their exact sum. Keeping the cancellation symbolic is both faster and less error-prone.

The governing concept is factorial cancellation. Expand the numerator only until the denominator factorial appears: (n + 3)! = (n + 3)(n + 2)(n + 1)!. Therefore, [(n + 3)(n + 2)(n + 1)!]/(n + 1)! simplifies by cancelling the common factor (n + 1)!, leaving (n + 3)(n + 2). Thus option A is correct. Option B incorrectly uses n + 1 as a separate factor instead of recognizing that the entire (n + 1)! cancels. Option C omits the factor n + 3, and option D retains the denominator factorial's factors unnecessarily. This method is valid wherever the factorials are defined, normally for nonnegative integer arguments.

The governing idea is cancellation in factorial notation. Since 14! = 14·13·12! and 2! = 2·1 = 2, the first term is 14·13/2 = 182/2 = 91. Similarly, 13! = 13·12·11!, so the second term is 13·12/2 = 156/2 = 78. Adding the two evaluated terms gives 91 + 78 = 169. Therefore option B is correct. A common error is to cancel only one factorial factor or to forget that 2! equals 2 rather than 1. Options A, C, and D result from incorrect cancellation or arithmetic, whereas 169 follows directly from expanding only the factors needed to cancel the denominators.

The governing concept is factorial cancellation, with the expressions also recognizable as binomial coefficients. For the first term, 10!/(6!·4!) = (10·9·8·7)/(4·3·2·1) = 5040/24 = 210, which is 10 choose 4. For the second term, 10!/(7!·3!) = (10·9·8)/(3·2·1) = 720/6 = 120, which is 10 choose 3. Therefore the required sum is 210 + 120 = 330. Option D is correct. The other choices do not result from the exact factorial cancellations; a frequent mistake is to cancel 10! incorrectly or to use 4! in the second denominator instead of 3!. Evaluating each term separately avoids that error.

Use the factorial identity m! = m(m − 1)(m − 2)!, expanding the numerator only until (3n − 1)! appears. We get (3n + 1)! = (3n + 1)(3n)(3n − 1)!. Hence [(3n + 1)(3n)(3n − 1)!]/(3n − 1)! cancels to (3n + 1)(3n). Therefore option A is correct. Option B incorrectly replaces the middle factor 3n with 3n − 1. Option C omits the first factor 3n + 1, and option D includes (3n − 1) as an extra factor even though the corresponding factorial has already been cancelled. The simplification assumes the factorials are defined, as in the usual natural-number domain for n.

Answer: 4, so option B is correct. Expand the numerator only until the denominator appears: (3n+1)!=(3n+1)(3n)(3n−1)!. After cancelling (3n−1)!, the equation becomes (3n+1)(3n)=156. Put t=3n. Then t(t+1)=156. The consecutive integers 12 and 13 have product 12×13=156, so t=12 and 3n=12. Dividing by 3 gives n=4. Verification: for n=4, the original expression is 13!/11!=13×12=156. Option A gives 10×9=90, option C gives 16×15=240, and option D gives 19×18=342. Thus only B works. The key is to identify the two consecutive factors rather than expand all factorials.

Answer: 11, so option B is correct. Simplify the factorials using n!=n(n−1)! and (n+1)!=(n+1)n!. The left side is [n(n−1)!·n!]/[(n−1)!(n+1)n!]. The common factors (n−1)! and n! cancel, leaving n/(n+1). Therefore n/(n+1)=11/12. Cross-multiply: 12n=11(n+1)=11n+11, so n=11. Substitution gives 11/12, confirming the result. Option A would give 10/11, option C would give 12/13, and option D would give 13/14; none equals 11/12. A useful warning is that n! is not n+n or n×n: it is the product 1×2×…×n. Writing neighboring factorials in terms of a common factorial makes the cancellation clear.

Answer: 7, so option B is correct. Expand (n+5)! down to (n+1)!: (n+5)!=(n+5)(n+4)(n+3)(n+2)(n+1)!. Cancelling the denominator gives (n+5)(n+4)(n+3)(n+2)=11880. Test the choices or match consecutive factors. For n=7, the factors are 12, 11, 10 and 9, and 12×11×10×9=132×90=11880. Thus n=7. Option A gives 11×10×9×8=7920, option C gives 13×12×11×10=17160, and option D gives 14×13×12×11, which is still larger. Since the product of positive consecutive factors increases as n increases, the matching choice is unique. Memory cue: cancel the smaller factorial first, then inspect the remaining consecutive factors.

The governing concept is expansion and cancellation of factorials. For the first quotient, (n+6)!/(n+3)! leaves the factors (n+6)(n+5)(n+4). For the second quotient, (n+5)!/(n+2)! leaves (n+5)(n+4)(n+3). Therefore the difference is (n+6)(n+5)(n+4)−(n+5)(n+4)(n+3). Factor out the common expression (n+5)(n+4), obtaining (n+5)(n+4)[(n+6)−(n+3)]. The bracket equals 3, so the result is 3(n+5)(n+4), making option B correct. Option A incorrectly uses a difference of 2, while C and D do not follow from the two remaining consecutive factors. The method avoids unnecessarily expanding the factorials into large products.

Use factorial expansion and cancel common factors carefully. The first fraction is (n+3)!/(n−2)! = (n+3)(n+2)(n+1)n(n−1). The second is (n+2)!/(n−3)! = (n+2)(n+1)n(n−1). Dividing the first by the second cancels the common factors (n+2)(n+1)n(n−1), leaving n+3, provided the factorials are defined. The corrected equation is therefore n+3 = 7, so n = 4. Hence option B is correct. The original right-hand side 5/2 would lead to n = −1/2, which is not among the positive integer options and is incompatible with the factorial notation used here.

Use factorial cancellation and expand each quotient only as far as needed. For the first term, (n+4)!/(n+1)! = (n+4)(n+3)(n+2). For the second term, (n+3)!/n! = (n+3)(n+2)(n+1). The common factor in both terms is (n+3)(n+2), so factor it out: (n+3)(n+2)[(n+4) + (n+1)]. Adding the remaining factors gives 2n + 5. Therefore the simplified form is (n+3)(n+2)(2n+5), which is option C. Option B incorrectly treats (n+4) as common to both terms, while A has the wrong final linear factor. Option D omits the correct factors from the first quotient.

The highest power of 10 dividing 40! is determined by counting complete factors of 10, and each factor of 10 contains one 2 and one 5. Since 40! contains many more factors of 2 than factors of 5, the number of available pairs is limited by the exponent of 5. Using Legendre’s formula, the exponent of 5 in 40! is floor(40/5) + floor(40/25) = 8 + 1 = 9. Therefore 40! is divisible by 10^9 but not by 10^10, so option B is correct. The value 8 omits the extra factor 5 from 25, while 10 and 11 require more factors of 5 than are present.

Answer: 7, so option A is correct. Expand the numerator to the denominator: (n+4)!=(n+4)(n+3)(n+2)(n+1)!. Cancelling (n+1)! gives (n+4)(n+3)(n+2)=990. The three factors are consecutive. Since 990=9×10×11, match n+2=9, n+3=10 and n+4=11. Therefore n=7. Check in the original expression: 11!/8!=11×10×9=990. Option B, n=8, gives 12×11×10=1320; option C gives 13×12×11=1716; and option D gives 14×13×12=2184. Thus none of them works. A useful cue is to arrange the remaining factors in increasing order; the smallest one is n+2, not n itself.