Question 1152304
R!/((R-3)!(R-4)!) = 1


Using trial and error...
8!/(5!*4!) = 8*7*6/(4!) = 336/24 > 1
9!/(6!*5!)  = 9*8*7/(5!) = 504/120 > 1
10!/(7!*6!) = 10*9*8/(6!) = 720 / 720 = 1


So <b>R = 10</b> satisifies the equation.
The sum of the digits of R is equal to {{{ highlight( 1 ) }}}