Factorial notation

Correct answer: B. (40320)

(8!=8\times7!) and (7!=5040), so the value is (40320). Find a larger factorial using the previous factorial.

Correct answer: B. 12

Since \(12!=12\times11!\), \(\frac{12!}{11!}=\frac{12\times11!}{11!}=12\). Option 132 is \(12\times11\), but the entire factor \(11!\) cancels in the division. Exam tip: remember that \(\frac{n!}{(n-1)!}=n\).

Correct answer: C. \(6\times5\times4!\)

By the definition of factorial, \(6!=6\times5\times4\times3\times2\times1\) and \(4!=4\times3\times2\times1\). Therefore, \(6!=6\times5\times4!\). In \(6\times4!\), the factor \(5\) is missing. Exam tip: When expressing \(n!\) in terms of \((n-r)!\), include every intervening factor.

Correct answer: A. 722

For factorials, 2! = 2 and 6! = 6×5×4×3×2×1 = 720. Therefore, 2! + 6! = 2 + 720 = 722. The option 720 is only the value of 6!; it does not include 2!. Exam tip: evaluate each factorial separately before performing the given operation.

Correct answer: B. (720)

The direct answer is B, 720. Use the factorial rule by expanding only the factors that do not cancel: \(10!=10\times9\times8\times7!\). Therefore, \(\frac{10!}{7!}=\frac{10\times9\times8\times7!}{7!}=10\times9\times8\). Calculate step by step: \(10\times9=90\), and \(90\times8=720\). Option A, 540, is wrong because it is not the product of the three remaining factors. Option B, 720, is correct because the common \(7!\) cancels exactly. Option C, 840, is wrong and would require a different product, not the factors left after cancellation. Option D, 900, is also wrong because it treats the expression as something other than \(10\times9\times8\). The useful rule is \(\frac{n!}{r!}=(r+1)(r+2)\cdots n\) when the numerator contains the denominator factorial.

Correct answer: C. \(6\)

Given \(n=5\), \((n-2)!=(5-2)!=3!=3\times2\times1=6\). Therefore, \(6\) is the correct option. The value \(24\) is \(4!\), not \(3!\). In exams, simplify the expression inside the brackets before applying the factorial.

Correct answer: B. 84

The governing concept is cancellation of common factorial factors. Expand only the part of 9! beyond 6!: 9! = 9 × 8 × 7 × 6!. Therefore, 9!/(6!·3!) = (9 × 8 × 7)/3!. Since 3! = 3 × 2 × 1 = 6, the value is (9 × 8 × 7)/6 = 504/6 = 84. Hence option B is correct. Option A can result from omitting one factor in the numerator. Option C reflects an incorrect division, while option D does not follow from the factorial cancellation. It is important to cancel 6! correctly and still evaluate the complete denominator factor 3!, not just the number 3.

Correct answer: C. 120

By definition, \(5!=5\times4\times3\times2\times1=120\) and \(0!=1\). Therefore, \(5!\times0!=120\times1=120\), so option C is correct. Option A would result from the incorrect assumption that \(0!=0\). Exam tip: always remember that \(0!=1\) because it is the value of an empty product.

Correct answer: C. (11!)

The direct answer is option C, 11!. Factorial notation is built by multiplying a number by the factorial immediately below it: \(11!=11\times10\times9!\). The given expression is already exactly this product, so its value is \(11!\). Option A, 9!, is too small because the factors 11 and 10 are still present. Option B, 10!, is also too small because \(10!=10\times9!\) and does not include the additional factor 11. Option C, 11!, is correct because its expansion is precisely \(11\times10\times9!\). Option D, 12!, is too large; it would be \(12\times11\times10\times9!\), containing one extra factor 12. The important skill is recognizing a factorial expansion without calculating the very large product. Exam cue: when consecutive descending factors end at 1 through a factorial, match the first number.

Correct answer: B. 5041

Here, \(7!=7\times6\times5\times4\times3\times2\times1=5040\), and by definition \(0!=1\). Therefore, \(7!+0!=5040+1=5041\). Option A is only the value of \(7!\); it does not include the 1 from \(0!\). Exam tip: Always remember that \(0!=1\).

Correct answer: B. 56

Since \(8!=8\times7\times6\times5!\), \(\frac{8!}{3!\,5!}=\frac{8\times7\times6}{3!}=\frac{8\times7\times6}{6}=56\). Hence, the correct answer is 56. Option 60 is a close distractor, but it does not divide correctly by \(3!=6\). Exam tip: Expand the larger factorial only up to the smaller factorial, then cancel common factors first.

Correct answer: C. 144

Here, 4! = 4 × 3 × 2 × 1 = 24 and 3! = 3 × 2 × 1 = 6. Therefore, xy = 24 × 6 = 144, so option C is correct. 132 is not the correct product of 24 and 6. Exam tip: Evaluate each factorial first, then substitute the values into the given expression.

Correct answer: B. 138

Using factorial values, \(4!=24\), \(5!=120\), and \(3!=6\). Therefore, \(4!+5!-3!=24+120-6=138\), so 138 is the correct option. A value such as 144 can result from an error in subtraction. Exam tip: evaluate each factorial separately before performing addition or subtraction.

Correct answer: C. \(n+5\)

Using factorial expansion, \((n+5)!=(n+5)(n+4)!\). Therefore, \(\frac{(n+5)!}{(n+4)!}=n+5\), since \((n+4)!\) cancels out. Choosing \(n+4\) misses the extra factor \(n+5\) in the numerator. Exam tip: Expand the larger factorial up to the smaller factorial, then cancel the common factor.

Correct answer: A. 720

Using factorial notation, \(3!=3\times2\times1=6\) and \(5!=5\times4\times3\times2\times1=120\). Therefore, \(3!\times5!=6\times120=720\), so option A is correct. Option B, 600, may result from incorrectly taking \(5!\) as 100; actually, \(5!=120\). Exam tip: evaluate each factorial separately before multiplying.

Correct answer: A. (\frac{7!}{4!})

(7!=7\times6\times5\times4!), so (\frac{7!}{4!}=7\times6\times5). Choose the denominator according to the remaining factors.

Correct answer: C. 7

Since \(6!=6\times5!\), \(\frac{6!+5!}{5!}=\frac{6\times5!+5!}{5!}=\frac{7\times5!}{5!}=7\). Option 6 would result if the numerator contained only \(6!\); here, \(5!\) is also added. Exam tip: Rather than evaluating factorials separately, factor out the smaller factorial to simplify quickly.

Correct answer: B. 6

Since \(6!=6\times5!\), \(\frac{a}{b}=\frac{6!}{5!}=\frac{6\times5!}{5!}=6\). Therefore, option B is correct. \(30\) is incorrect because it is the value of \(5!\), not of \(\frac{6!}{5!}\). Exam tip: use the direct rule \(\frac{n!}{(n-1)!}=n\).

Correct answer: B. 35280

Since \(8!=8\times7!\), \(8!-7!=8\times7!-7!=7!(8-1)=7\times7!\). As \(7!=5040\), the value is \(7\times5040=35280\). \(40320\) is only the value of \(8!\); it does not account for subtracting \(7!\). Exam tip: For consecutive factorials, factor out the smaller factorial first.

Correct answer: B. 26

First evaluate the bracket: \((5-1)! = 4! = 4\times3\times2\times1 = 24\). Also, \(2! = 2\times1 = 2\). Therefore, \((5-1)!+2! = 24+2=26\). Option 24 is only the value of \(4!\); it misses the addition of \(2!\). Exam tip: simplify the expression inside brackets before applying factorial.

Correct answer: C. 9

By definition, \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\). There are 9 numbers being multiplied, so it has 9 factors in its factorial expansion. \(8\) would be the number of factors in \(8!\), not \(9!\). Exam tip: In the expanded form of \(n!\), there are \(n\) factors from \(n\) to 1.

Correct answer: B. 11

Using factorials, \(5!=5\times4!\), so \(5!\div4!=5\). Also, \(3!=3\times2\times1=6\). Therefore, \(5!\div4!+3!=5+6=11\), so option B is correct. \(12\) may result from incorrectly evaluating \(5!\div4!\) as 6. Exam tip: simplify factorial expressions first, then perform division and addition.

Correct answer: C. \(13\times12\times11!\)

By the definition of factorial, \(13!=13\times12\times11\times10\times\cdots\times1\). Since \(11!=11\times10\times\cdots\times1\), we get \(13!=13\times12\times11!\). In \(13\times11!\), the factor \(12\) is missing. Exam tip: Expand the larger factorial only up to the given smaller factorial.

Correct answer: B. (1)

The factorial identity \(7!=7\times6!\) allows the numerator to be rewritten in exactly the same form as the denominator. The denominator is \(7\times6!\), so after expansion both numerator and denominator are equal. Since factorials of positive integers are non-zero, cancelling the common factor is valid and leaves a quotient of one equal quantity by itself.

Rewrite the numerator: \(7!=7\times6!\). Hence \(\frac{7!}{7\times6!}=\frac{7\times6!}{7\times6!}=1\). Equivalently, cancel \(7\) and \(6!\) from the numerator and denominator. Therefore option B, \(1\), is correct. The answer is not \(7\) or \(6\), because those factors cancel rather than remain in the quotient.

Correct answer: C. 26

Here, \(1!=1\), \(4!=4\times3\times2\times1=24\), and \(0!=1\). Therefore, \(1!+4!+0!=1+24+1=26\), so option C is correct. The value 25 results from omitting \(0!\), but \(0!\) is always 1. Exam tip: Remember the special values \(0!=1\) and \(1!=1\).