Question 166109

Since order does not matter (ie the seats have no unique positions), we must use the <a href=http://www.mathwords.com/c/combination_formula.htm>combination formula</a>:



*[Tex \LARGE \textrm{_{n}C_{r}=]{{{n!/(n-r)!r!}}} Start with the combination formula.




*[Tex \LARGE \textrm{_{6}C_{4}=]{{{6!/(6-4)!4!}}} Plug in {{{n=6}}} and {{{r=4}}}




*[Tex \LARGE \textrm{_{6}C_{4}=]{{{6!/2!4!}}}  Subtract {{{6-4}}} to get 2



Expand 6!
*[Tex \LARGE \textrm{_{6}C_{4}=]{{{(6*5*4*3*2*1)/2!4!}}}



Expand 2!
*[Tex \LARGE \textrm{_{6}C_{4}=]{{{(6*5*4*3*2*1)/(2*1)4!}}}




*[Tex \LARGE \textrm{_{6}C_{4}=]{{{(6*5*4*3*cross(2*1))/(cross(2*1))4!}}}  Cancel




*[Tex \LARGE \textrm{_{6}C_{4}=]{{{(6*5*4*3)/4!}}}  Simplify



Expand 4!
*[Tex \LARGE \textrm{_{6}C_{4}=]{{{(6*5*4*3)/(4*3*2*1)}}}




*[Tex \LARGE \textrm{_{6}C_{4}=]{{{360/(4*3*2*1)}}}  Multiply 6*5*4*3 to get 360




*[Tex \LARGE \textrm{_{6}C_{4}=]{{{360/24}}} Multiply 4*3*2*1 to get 24




*[Tex \LARGE \textrm{_{6}C_{4}=]{{{15}}} Now divide




So 6 choose 4 (where order doesn't matter) yields 15 unique combinations. So there are 15 possible ways that 6 people can sit in 4 chairs.