Question 1152304
<pre>                   
{{{R!/((R-3)!(R-4)!)=1}}}

R has to be 4 or more, so we start trying to find R, so we will get 1

{{{4!/((4-3)!(4-4)!)=4!/(1!0!)=24/(1*1)=24}}}

{{{5!/((5-3)!(5-4)!)=5!/(2!1!)=120/(2*1)=60}}}

{{{6!/((6-3)!(6-4)!)=6!/(3!2!)=720/(6*2)=60}}}

{{{7!/((7-3)!(7-4)!)=7!/(4!3!)=120/(24*6)=35}}}

{{{8!/((8-3)!(8-4)!)=8!/(5!4!)=40320/(120*24)=14}}}

{{{9!/((9-3)!(9-4)!)=9!/(6!5!)=362880/(720*120)=4.2}}}

{{{10!/((10-3)!(10-4)!)=10!/(7!6!)=3628800/(5040*720)=1}}}

Finally found it!  R=10

The sum of the digits of 10 is 1+0 = 1

Edwin</pre>