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If ( \frac{n!}{(n-3)!}=504 ), what is the positive value of (n)?

Answer and explanation

Correct answer: (9)

Direct answer: Option C, \\(n=9\\). Expand the factorial ratio by cancelling \\( (n-3)!\\): \\(n!/(n-3)!=n(n-1)(n-2)\\). Therefore solve \\(n(n-1)(n-2)=504\\). Testing consecutive factors, \\(9\cdot8\cdot7=72\cdot7=504\\), so n=9. Option A, n=7, gives \\(7\cdot6\cdot5=210\\), not 504. Option B, n=8, gives \\(8\cdot7\cdot6=336\\), not 504. Option C gives exactly 504 and is positive. Option D, n=10, gives \\(10\cdot9\cdot8=720\\), too large. Factorial ratios are products of consecutive descending integers, so looking near the cube root of 504 is useful; \\(8^3=512\\), which points to 9, 8, 7. Memory cue: cancel factorials before calculating.

Related tags

PermutationsCombinationsFactorialHard

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