Question 754301
In this situation, the "order" in which people are chosen to make up a group does not matter. (A situation where order would matter would be, for example, one where three people are chosen to fill three different offices or positions.)

Since order doesn't matter, you are dealing with a combination.

A combination of "n" people or objects, chosen "r" at a time would be given by:


n!/[r!(n-r)!]


In this case our "n" is 10, and "r" is 3. If you plug those into the formula you have:


10!/3!(10-3)!   =    10!/3!7!



= (10 X 9 X 8 X 7!)/[(3 X 2 X 1)(7!)]



The 7!'s cancel, so you have:



(10 X 9 X 8)/(3 X 2 X 1) , which simplifies to 720/6  or  120 possible
 combinations of sales representatives.