Question 1168882
<font face="Times New Roman" size="+2">



There are 4 guests and the left-most seat must be occupied, so there are 4 ways to choose the occupant of that seat.  Then, for each of those 4 ways, there are 4 ways to decide the occupant of the second seat -- the three remaining people plus the possibility of that seat being empty, for a total of 16 ways to decide the configuration of the first two seats.  Then, for each of those 16 ways, there are three ways to choose the occupant of the third seat (either 3 remaining guests if the second seat was left empty, or the two remaining guests plus the possibility of leaving the 3rd seat empty).  16 times 3 is 48.  For each of those ways, there are two choices for the next seat, so 96, and then one choice for the fifth seat.  So 96 ways.


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish]


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>