Factorial notation

Answer: 4(n+5)(n+4)(n+3), so option A is correct. Expand each factorial ratio only as far as its denominator. The first ratio is (n+6)!/(n+2)!=(n+6)(n+5)(n+4)(n+3). The second is (n+5)!/(n+1)!=(n+5)(n+4)(n+3)(n+2). The common factor is (n+5)(n+4)(n+3). Factoring it out leaves [(n+6)−(n+2)](n+5)(n+4)(n+3)=4(n+5)(n+4)(n+3). Option B uses the incorrect difference 3, and option D uses 2. Option C changes the common-factor structure and is not equivalent. The important habit is to align the products before subtracting.

The governing concept is the prime-exponent method for factorials, also called Legendre’s formula. We need v₂(n!) ≥ 9, v₃(n!) ≥ 3, and v₅(n!) ≥ 2. For 11!, v₂(11!) = ⌊11/2⌋ + ⌊11/4⌋ + ⌊11/8⌋ = 5 + 2 + 1 = 8, so the required power 2^9 is still missing. Also, v₃(11!) = 3 + 1 = 4 and v₅(11!) = 2, so those conditions are already met. On multiplying by 12, the factor 12 contributes two more powers of 2, giving v₂(12!) = 10. Thus 12! satisfies every condition, while 11! does not. Therefore option D, 12, is correct; 9, 10, and 11 fail the power-of-2 requirement.

The factorial-cancellation rule gives (n+4)!/(n−1)! = n(n+1)(n+2)(n+3)(n+4). Dividing by (n+3)(n+2) leaves n(n+1)(n+4). The corrected equation is therefore n(n+1)(n+4) = 1170. Testing the available positive integer values, n = 9 gives 9 × 10 × 13 = 1170. Hence n = 9, so option B is correct. The nearby values do not work: n = 8 gives 8 × 9 × 12 = 864, n = 10 gives 10 × 11 × 14 = 1540, and n = 11 gives 11 × 12 × 15 = 1980. The original right-hand side 990 had no matching listed integer option, so it was corrected to preserve a valid MCQ.