Question 773503
He can have 1 friend for dinner
He can have 2 friends for dinner
He can have 3 friends for dinner 
and so on up to
He can have 8 friends for dinner
----------------------------
In each of these 8 cases, use the
formula for {{{ C(8,r) }}} where {{{ r }}} is 
the number of friends invited 
to dinner. Note that you don't use
permutations since the order of
the invited friends doesn't matter
----------------------------
So, the solution is:
{{{ C(8,1) + C(8,2) + C(8,3) + C(8,4) }}}
etc. up to {{{ C(8,8) }}}
-----------------
{{{ C(8,1) = 8! / ( 1!*( 8-1 )! ) }}}
{{{ C(8,1) = 8!/7! }}}
{{{ C(8,1) = 8 }}}
-----------------
{{{ C(8,2) = 8! / ( 2!*( 8 - 2 )! ) }}}
{{{ C(8,2) = 8! / ( 2*1*6*5*4*3*2*1 )}}}
{{{ C(8,2) = ( 8*7 ) / 2 }}}
{{{ C(8,2) = 28 }}}
---------------
Continue up to {{{ C(8,8) }}} and add up
all the results.