There are 10 books on a shelf, of which 4 are paperbacks and 6 are har

You can calculate directly possible selections of 5 books from the shelf that contain at least one paperback and at least one hardback directly as shyind did or indirectly as chetan2u did.

I will elaborate further on chetan2u's method. To calculate the possible selections of 5 books from the shelf that contain at least one paperback and at least one hardback, subtract the complement of at least one paperback and at least one hardback from total number of 5 book combinations. To find the complement event, use DeMorgan's rule.


DeMorgan's Rule states that ~(P and Q) = ~P or ~Q
~ symbol means Not

So if we let P = at least one paperback then ~P = No paperbacks.
So if we let Q = at least one hardback then ~Q = No hardbacks.

No hardbacks would mean 5 paperbacks, but this is not possible since there are only 4 paperbacks. ~Q = 0 ways
No paperbacks would mean 5 hardbacks, which can be chosen in \(C^6_5\) in 6 ways. ~P = 6 ways
~P or ~Q = ~P + ~Q = 6 + 0 = 6 ways

Total number of 5 book combinations from 10 books is \(C^{10}_5\) = \(\frac{(10*9*8*7*6)}{(5*4*3*2*1)}\) = 252 ways

(Total number of 5 book combinations) - (No paperbacks or No hardbacks) = 252 - 6 = 246.

Correct answer choice is D.