What is the value of \(\frac{9!-7!}{7!}\)?

Correct answer: 71

Since \(9!=9\times8\times7!\), \(\frac{9!}{7!}=9\times8=72\). Therefore, \(\frac{9!-7!}{7!}=\frac{9!}{7!}-\frac{7!}{7!}=72-1=71\). Option 72 is only the value of \(\frac{9!}{7!}\); subtracting \(\frac{7!}{7!}=1\) is still required. Exam tip: In factorial fractions, expand the larger factorial up to the smaller factorial before cancelling common factors.