Question 473736
This will be solved using combinations because there is no order within the sub-committees. 
How many groups of 3 can be made from 9 members.
Combination formula:
9C3 = {{{9!/((9-3)!*3!) = 9!/(6!*3!)}}} 
9! can be written as 9*8*7*6!
Notice that the 6! can be cancelled on top and bottom, leaving
{{{(9*8*7)/(3*2*1)}}}
You can simplify things by doing more cancelling, {{{9/3 = 3}}}, {{{8/2 = 4}}}
={{{3*4*7 = 12*7 = 84}}}
Therefore there are 84 sub-committees possible.