Question 1033270
THE FIFTH GRADER'S THOUGHT:
Anthony lets luck of the draw decide.
He writes the name of each college on a small piece of paper, and
puts the pieces of paper in a bag to draw the names at random.
He draws the first piece of paper, and marks it as the first college where he will spend a night.
Of course, he does not put that piece of paper back in the bag, before drawing again to find another college where he will spend a night.
He then draws {"without replacement") a second and a third piece of paper that decide where he will spend the second and third nights respectively.
That decides the schedule of college sleepovers.
There are {{{10*9*8=720}}} possible schedules
(10 possibilities for the first draw, 9 for the second, and 8 for the third draw).
However, each set of {{{3}}} pieces of papers could be drawn in
{{{3*2*1=6}}} orders, and it would still be the same set, just different orders.
There are {{{720}}} different college sleepover schedules,
but they represent {{{720/6=120}}} different sets of sleepover college choices.
The chances that {{{1}}} particular set of colleges could be chosen that way is {{{1}}} in {{{120}}} ,
so the probability that Anthony spends a night at Rutgers University, a night at the University of Miami, and a night at Clemson University is
{{{1/120}}} .
 
THE MATH TEACHER SAYS:
This is a case of combinations (setof items, where order does not matter).
The number of combinations of {{{r}}} items taken without replacement from a set of {{{n}}} items is
{{{(matrix(2,1,n,r))=n!/(r!(n-r)!)}}} .
In this case {{{n=10}}} and {{{r=3}}} ,
so the number of possible sets of {{{3}}} colleges where Anthony could spend the night,
out of the {{{10}}} colleges he wanted to visit is
{{{(matrix(2,1,10,3))=10!/(3!(10-3)!)=10!/(3!7!)=10*9*8*7*6*5*4*3*2*1/(3*2*1)(7*6*5*4*3*2*1))=10*9*8*cross(7)*cross(6)*cross(5)*cross(4)*cross(3)*cross(2)*cross(1)/(3*2*1*cross(7)*cross(6)*cross(5)*cross(4)*cross(3)*cross(2)*cross(1)))=720/6=120}}} .
{Rutgers U, U Miami, Clemson U} is one 3-college subset of the 10-college set of colleges that Anthony wants to visit.
The probability that the set of 3 colleges where Anthony spends a night
is that {{{1}}} particular set out of the {{{120}}} possible sets is
{{{1/120}}} .