At a family party, 18 people need to be seated at two tables, a table for eight and a table for ten. In how many ways can the table assignments be made ?
Thank you !
Don't be confused by there being two tables - once you have chosen the people to sit at one table, the other is determined. Everyone not picked for one table has to sit at the other.
So your answer is the number of combinations of 18 possible things taken 8 at a time.
The formula for combinations is:
$\displaystyle C(n,k)=\frac{n!}{k!(n-k)!}$
There are a variety of notations used (where I used C(n,k)). Note that the formula is symmetric - C(n,k) = C(n,n-k). That is consistent with our problem - we can choose either 8 people for the first table or 10 people for the second table, and the number of combinations has to be the same.
Post again if you're still having trouble.