Key observation: It's impossible to have a selection of 5 books, in which none of the books are hardbacks.
In other words, there will always be at least 1 hardback book in a collection of 5 books, which means we just have to deal with having at least one paperback

Well use to formula: # of ways to obey the restriction = (# of ways to ignore the restriction) - (# of ways to break the restriction)

# of ways to ignore the restriction
In other words, in how many ways can we select 5 books from 10 books?
Since the order in which we select the books does not matter, we can use combinations.
We can select 5 books from 10 books in 10C5 ways
\(10C5 = \frac{(10)(9)(8)(7)(6)}{(5)(4)(3)(2)(1)} = \frac{(9)(8)(7)(6)}{(4)(3)(1)}= \frac{(9)(2)(7)(6)}{(3)(1)}= (3)(2)(7)(6) = 252\)

# of ways to break the restriction
In order to break the restriction, we must have 0 paperbacks in the selection of 5 books,
In other words, we must select 5 books from the 6 hardbacks.
Once again we'll use combinations.
We can select 5 hardbacks from 6 hardbacks in 6C5 ways
\(6C5 = \frac{(6)(5)(4)(3)(2)}{(5)(4)(3)(2)(1)} = 6\)

So, # of ways to obey the restriction \(= 252 - 6 = 246\)

Answer: D