An article quotes that "there are more than 1,000 ways to assign ten customers to two salespeople" however this is wrong! How many actual ways are there to divide these customers across the two salespeople and what is the method used to get to this number?
I believe I should use combinations to answer this question but i'm not sure if im right. My thoughts were if the combination formula is nCr = n!/(n−r)!r! then it would be 10c5 = 10!/((10-5)!5! = 252
Thank you for any help you can provide :)
On
The customers, being humans, are distinct. There is no restriction on splitting the load, so as stated the quote is right; each customer may go to either salesperson independently, and there are $2^{10}=1024$ choices overall.
You assume in deriving your answer of $\binom{10}5=252$ that the load must be distributed evenly and that order doesn't matter.
If each customer can be assigned to either salesperson, the answer is $2^{10}=1024$. That is clearly what the statement is thinking. Each customer can be assigned $2$ ways and you use the multiplication principle to combine them. You are assuming that each salesperson must be responsible for $5$ customers, but that is not stated in the problem.