Question 37691
To do this problem you need Combinations and Permutations.
You know for permutations, the order of the object matter.  In this case, the order doesn't matter, so we will be using combinations. 

{{{nCr}}}
{{{n!/((n-r)!*r!)}}}

n equals the number of choices.  In this case, you have 10.
r equals the number of decisions yo make.  In this case, you have 3.
Plug your numbers in!

{{{10!/((10-3)!*3!)}}}
Simplify 10! with 7!, you will be left with {{{10*9*8}}}
That equals 720

{{{720/3!}}}
{{{720/6}}}
Answer:{{{highlight(120)}}}

Hope this helps!