Question 199877
Unfortunately, the previous solution is incorrect. Why? They're not looking for a probability, they just want to know the number of possible combinations.





We could use the counting principle to solve this problem, but we'll have overlap and the sample space is far too large. So let's do it this way:



In this case, order does NOT matter since the candidates have no rank over one another (ie one isn't president or secretary). 



Since order does not matter, 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{_{15}C_{6}=]{{{15!/(15-6)!6!}}} Plug in {{{n=15}}} and {{{r=6}}}



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{15!/9!6!}}}  Subtract {{{15-6}}} to get 9



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{(15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/9!6!}}} Expand 15!



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{(15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/(9*8*7*6*5*4*3*2*1)6!}}} Expand 9!



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{(15*14*13*12*11*10*cross(9*8*7*6*5*4*3*2*1))/(cross(9*8*7*6*5*4*3*2*1))6!}}}  Cancel



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{(15*14*13*12*11*10)/6!}}}  Simplify



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{(15*14*13*12*11*10)/(6*5*4*3*2*1)}}} Expand 6!



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{3603600/(6*5*4*3*2*1)}}}  Multiply 15*14*13*12*11*10 to get 3,603,600



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{3603600/720}}} Multiply 6*5*4*3*2*1 to get 720



*[Tex \LARGE \textrm{_{15}C_{6}=]{{{5005}}} Reduce.



So 15 choose 6 (where order doesn't matter) yields 5,005 unique combinations



This means that there are 5,005  different ways to select a group of 6 college candidates from a group of 15 applicants for an interview (where the order of the candidates doesn't matter).