MathRevolution wrote:
[GMAT math practice question]
If \(\frac{n!}{(n-2)!}<100\), what is the greatest possible value of \(n\)?
\(A. 8\)
\(B. 9\)
\(C. 10\)
\(D. 11\)
\(E. 12\)
i would like to use back solving strategy. !
but before doing so , a simple thing that must be mentioned is that numerator is n! and denominator is (n-2)!
it means that 2 no. will always be extra in numerator part.
how?
a) 8 as a value of n
so, 8!/ (8-2)!
= 8!/6!
= 8*7*6*5*4*3*2*1/6*5*4*3*2*1
= 8*7
=56 <100 , it meets our condition .
now , we have to remember that we are looking for max value of n
it must be 10
10!/(10-2)!
=10*9*8*7*6*5*4*3*2*1/8*7*6*5*4*3*2*1
=10*9
=90 <100
the correct answer is C.