Factorial notation

(7!=7\times6\times5\times4\times3\times2\times1=5040). Find a larger factorial by multiplying the previous factorial.

By the definition of factorial, \(8!=8\times7\times6!\). Therefore, \(8\times7\times6!\) is correct. In \(8\times6!\), the factor \(7\) is missing, so it is not equal to \(8!\). Exam tip: When expressing \(n!\) in terms of \((n-r)!\), include every consecutive factor in between.

Since \(9!=9\times8\times7!\), \(\frac{9!}{7!}=\frac{9\times8\times7!}{7!}=9\times8=72\). Option C, \(90\), could result from incorrectly using \(9\times10\). In exams, expand a factorial only up to the factorial in the denominator so that it cancels directly.

Using factorial notation, \(6!=6\times5\times4\times3\times2\times1=720\) and \(1!=1\). Therefore, \(6!+1!=720+1=721\). Option 719 would result from subtracting 1 from 720, but the question requires addition. Exam tip: evaluate each factorial first, then apply the indicated operation.

Since \(8! = 8\times7\times6!\), \(\frac{8!}{6!\,2!}=\frac{8\times7\times6!}{6!\times2}=\frac{56}{2}=28\). Therefore, 28 is correct. The close distractor 56 results from forgetting to divide by \(2!\). Exam tip: cancel the common factorial first, then multiply the remaining terms.

Here n=4, so n! = 4! = 4 × 3 × 2 × 1 = 24. Therefore, 24 is the correct option. The value 16 is the square of 4, not its factorial. Exam tip: To evaluate n! for a positive integer, multiply all whole numbers from n down to 1.

Using factorial values, 2! = 2 and 4! = 4 × 3 × 2 × 1 = 24. Also, 1! = 1. Therefore, 2! + 4! - 1! = 2 + 24 - 1 = 25, so option C is correct. Choosing 24 would ignore the effect of 2! and 1!. Exam tip: evaluate each factorial first, then perform addition or subtraction.

The direct answer is option B, 10. Factorial means multiplying all positive integers down to 1: for example, 10! = 10 × 9 × 8 × ... × 1. The useful fact is 10! = 10 × 9!, because 9! already contains 9 × 8 × ... × 1. Therefore, \(\frac{10!}{9!}=\frac{10\times9!}{9!}=10\), since the non-zero factor 9! cancels. Option A, 9, is wrong because the extra factor in 10! is 10, not 9. Option B, 10, is correct because exactly 10 remains after cancellation. Option C, 19, incorrectly adds 10 and 9 instead of dividing factorials. Option D, 90, incorrectly multiplies 10 by 9 and forgets that 9! is cancelled. Memory cue: \(\frac{n!}{(n-1)!}=n\).

Using factorial notation, \(4!=4\times3\times2\times1=24\) and \(3!=3\times2\times1=6\). Therefore, \(4!\times3!=24\times6=144\), so option C is correct. \(120\) is the value of \(5!\), not of this product. Exam tip: Evaluate each factorial first, then multiply the results.

(6!=6\times5\times4\times3!), so the given form is (6!). Recognizing expansions is important in factorial questions.

Since \(5!=5\times4!\), \(\frac{5!+4!}{4!}=\frac{5\times4!+4!}{4!}=\frac{6\times4!}{4!}=6\). Therefore, the correct answer is 6. Option 5 would result if the additional \(4!\) in the numerator were ignored. Exam tip: In factorial fractions, factor out the common factorial before simplifying.

By the definition of factorial, \(6!=6\times5\times4\times3\times2\times1=720\). Therefore, \(n=6\). Since \(5!=120\) and \(7!=5040\), those values cannot satisfy the condition. Exam tip: Memorising factorial values from \(1!\) to \(7!\) helps solve such questions quickly.

First evaluate the bracket: \(2+3=5\). Therefore, \((2+3)! = 5! = 5\times4\times3\times2\times1 = 120\), so option C is correct. \(24\) is the value of \(4!\), whereas \(720\) is the value of \(6!\). Exam tip: Always simplify the expression inside brackets before applying factorial.

Use the factorial recurrence 7! = 7 × 6! and factor out the common quantity 6!. The expression becomes 7! − 6! = 7 × 6! − 6! = (7 − 1)6! = 6 × 6!. Since 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720, the result is 6 × 720 = 4320. Therefore, option A is correct. A direct check also gives 7! = 5040 and 5040 − 720 = 4320. Option B is just 7!, while option C is just 6!, so neither is the requested difference. Option D results from an arithmetic or factoring error. Extracting the common factorial is the most efficient method.

Since \(6!=6\times5!\), \(\frac{6!}{5!}=6\). Also, \(2!=2\times1=2\). Therefore, \(\frac{6!}{5!}+2!=6+2=8\), so option C is correct. Option 7 may result from incorrectly taking \(2!\) as 1. Exam tip: simplify factorial fractions first by cancelling the common factorial term.

By the definition of factorial, \(3!=3\times2\times1=6\). Therefore, \(3!\times3!=6\times6=36\), so option D is correct. The value \(24\) equals \(4!\), not \(3!\times3!\). Exam tip: evaluate each factorial first, then perform the required operation.

Since \(7!=7\times6\times5\times4!\), \(\frac{7!}{4!}=7\times6\times5=210\). The value \(120\) is only \(5!\); here, the factors 7, 6, and 5 must be multiplied. Exam tip: In \(\frac{n!}{r!}\), cancel \(r!\) and multiply the integers from \(r+1\) to \(n\).

Using factorial values, 1! = 1, 2! = 2, and 3! = 6. Therefore, 1! + 2! + 3! = 1 + 2 + 6 = 9. Option 8 is incorrect because it is one less than the correct total after evaluating 3! correctly. Exam tip: evaluate each factorial separately before adding.

Using factorial expansion, \((n+3)!=(n+3)(n+2)!\). Therefore, \(\frac{(n+3)!}{(n+2)!}=n+3\), because \((n+2)!\) cancels from the numerator and denominator. \(n+2\) is a common error caused by omitting the extra factor \(n+3\). Exam tip: Expand the larger factorial only up to the smaller factorial, then cancel common factors.

By factorial expansion, \(8!=8\times7\times6\times5!\). Hence, the given expression equals \(8!\). In contrast, \(7!=7\times6\times5!\), so it does not include the factor 8. Exam tip: Match consecutive descending factors with the factorial at the end to identify the required factorial.

Here, \(4!=4\times3\times2\times1=24\) and \(2!=2\times1=2\). Therefore, \(\frac{4!+2!}{2!}=\frac{24+2}{2}=\frac{26}{2}=13\). Hence, 13 is the correct option. Option 12 can result from incorrectly evaluating \(2!\) or mishandling the addition in the numerator. Exam tip: evaluate each factorial first, add the terms in the numerator, and then divide.

\(5!\div3!=\frac{5\times4\times3!}{3!}\). The common factor \(3!\) cancels from the numerator and denominator, leaving \(5\times4=20\). Choosing 15 would result from an incorrect cancellation of the factorial terms. Exam tip: In factorial division, expand the larger factorial only up to the smaller factorial and cancel common factors.

Here, a=3!=3×2×1=6 and b=2!=2×1=2. Therefore, a+b=6+2=8, so option C is correct. The value 6 is only the value of a, not their sum. Exam tip: Evaluate each factorial separately before performing the required operation.

Since \(9!=9\times8\times7\times6!\), \(\frac{9!}{6!}=9\times8\times7=504\). The value \(3024\) would result from multiplying by an extra \(6\), but the entire \(6!\) cancels here, leaving only \(9,8,7\). Exam tip: In \(\frac{n!}{r!}\), where \(n>r\), expand only up to \(r!\) and cancel it.

By definition, \(0!=1\), and \(4!=4\times3\times2\times1=24\). Therefore, \(0!+4!=1+24=25\). Option 24 is only the value of \(4!\); it omits \(0!\). Exam tip: always remember the special value \(0!=1\).