Factorial notation

The governing concept is factorial notation and cancellation. For a positive integer n, n! is the product of all positive integers from n down to 1. Since 5! = 5 × 4!, the fraction 5!/4! simplifies directly to 5. Therefore, (5! × 3!)/4! = 5 × 3!. Now calculate 3! = 3 × 2 × 1 = 6, giving 5 × 6 = 30. Hence option B is correct. A direct check also confirms the result: 5! = 120, 3! = 6, and 4! = 24, so 120 × 6 ÷ 24 = 30. The factorial symbols must be evaluated as products, not treated as ordinary multiplication signs or cancelled without preserving their factors.

The governing concept is factorial cancellation followed by ordinary addition. Since 7! = 7 × 6 × 5!, the quotient 7!/5! simplifies to 7 × 6 = 42. Separately, 3! = 3 × 2 × 1 = 6. Therefore the value of the complete expression is 42 + 6 = 48, so option C is correct. Option B gives only the quotient and ignores the added 3!. Option A may result from an incorrect reduction or subtraction, while option D uses an incorrect value for 3! or makes an addition error. The efficient method is to cancel 5! first instead of expanding 7! into 5040; this illustrates why factorial notation is useful in calculations.

The governing idea is factoring a common factorial. Because 8! = 8 × 7!, the numerator becomes 8 × 7! − 7! = (8−1) × 7! = 7 × 7!. Dividing by the denominator 7! gives (7 × 7!)/7! = 7. Therefore option B is correct. Another valid approach is to divide the two numerator terms separately: 8!/7! − 7!/7! = 8 − 1 = 7. A common error is to treat 8! − 7! as 1!, but factorial values cannot be subtracted by subtracting their arguments. Option C forgets the subtraction of 7!, option D has no valid factorial derivation, and option A is an incorrect reduction.

The relevant concept is factorial simplification together with the order of operations. Factorials are evaluated within their own terms before subtraction. Since 6! = 6 × 5 × 4!, the quotient 6!/4! simplifies to 6 × 5 = 30. Also, 2! = 2 × 1 = 2. Consequently, 6!/4! − 2! = 30 − 2 = 28. Therefore option B is correct. Option C is the quotient before subtracting 2!, so it represents an incomplete calculation. Options A and D cannot be obtained from the stated expression. Cancelling 4! is legitimate because it is a complete factor in 6!, but the subtraction must be performed only after the quotient has been simplified.

Use the adjacent-factorial identity n!/(n−1)! = n, which follows directly from n! = n × (n−1)!. Thus 9!/8! = 9 because 9! = 9 × 8!, and 6!/5! = 6 because 6! = 6 × 5!. Adding these simplified values gives 9 + 6 = 15. Therefore, option C is correct. It is unnecessary and inefficient to calculate the very large values of 9! and 6! separately. Option A is two less than the correct sum, option B can result from an arithmetic mistake, and option D is one more than the correct answer. The denominators 8! and 5! are non-zero, so the cancellation is valid and the result is exact.

Use the factorial expansion rule and stop when the denominator factorial appears. The numerator is (n+6)! = (n+6)(n+5)(n+4)(n+3)!. Dividing by (n+3)! cancels the final factorial completely, leaving (n+6)(n+5)(n+4). Therefore option B is correct. There are three steps between n+6 and n+3, so exactly three factors remain. Option A omits n+4, option C includes n+3 even though that factor belongs to the denominator factorial, and option D skips n+5. Assuming the factorials are defined, this method is valid and avoids calculating very large factorial values directly.

The governing concept is cancellation in factorial notation. Since the denominator contains 18!, express only the needed part of the numerator: 20! = 20×19×18!. Substitute this into the fraction: 20!/(18!×19) = (20×19×18!)/(18!×19). The factor 18! cancels with 18!, and the factor 19 cancels with the denominator’s 19. The remaining value is 20, so option C is correct. A frequent mistake is to write 20! as 20×18!, omitting the factor 19; that invalid step can lead to an incorrect answer. Another mistake is to cancel only 18! and then stop before dividing by 19. The factorial identity confirms the result exactly, without calculating the enormous value of 20!.

The governing concept is the factorial identity k! = k(k−1)! and cancellation of common nonzero factors. In (7!/5!)×(5!/6!), the 5! in the numerator of the second fraction cancels the 5! in the denominator of the first fraction. The expression therefore becomes 7!/6!. Using 7!=7×6!, we obtain 7!/6!=7. A separate check gives 7!/5!=7×6=42 and 5!/6!=1/6, so their product is 42/6=7. Hence option C is correct. The value 6 results from retaining only one factor from the first ratio, while 8 would reflect an incorrect factorial step. Careful cancellation preserves the multiplication structure and confirms that the answer is uniquely 7.

Answer: 8, so option C is correct. Simplify each factorial ratio: 8!/6!=8×7 because 8!=8×7×6!, while 7!/6!=7 because 7!=7×6!. The expression therefore becomes (8×7)÷7=8. Another method confirms it: division by a fraction means multiplication by its reciprocal, so (8!/6!)÷(7!/6!)=(8!/6!)×(6!/7!)=8!/7!=8. Option A, 6, leaves the denominator factorial without completing the cancellation; option B, 7, simplifies only the divisor; and option D, 9, has no supporting factorial calculation. Do not cancel symbols across a division sign carelessly. First rewrite the division as multiplication by the reciprocal or simplify each ratio separately.

For a positive integer n, n! = n × (n−1) × ... × 1. Thus, 2! = 2 × 1 = 2. Therefore, 2! + 2! + 2! = 2 + 2 + 2 = 6. Option 8 is incorrect because each 2! equals 2, not 4. Exam tip: Evaluate each factorial separately before performing the given operations.