swerve wrote:There are 8 books in a shelf that consists of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?

A. 40
B. 45
C. 50
D. 55
E. 60

The OA is D.

Please, can anyone explain this PS question? I tried to solve it but I can't get the correct answer. I need help. Thanks.

Combinations with at least 1 paperback = (all possible combinations) - (combinations with no paperbacks).

All possible combinations:
From the 8 books, the number of ways to choose 4 = 8C4 = (8*7*6*5)/(4*3*2*1) = 70.

Combinations with no paperbacks:
From the 6 hardback books, the number of ways to choose 4 = 6C4 = (6*5*4*3)/(4*3*2*1) = 15.

Combinations with at least 1 paperback:
70-15 = 55.

The correct answer is D.

swerve wrote:There are 8 books in a shelf that consists of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?

A. 40
B. 45
C. 50
D. 55
E. 60

We can use the equation:

It must be true that:

The number of ways in which at least 1 paperback book is selected = The total number of ways to select 4 books - The number of ways in which no paperback books are selected

The number of ways in which no paperback books are selected is equivalent to the number of ways in which all 4 books selected are hardcover. Let's determine that now. There are 6 hardback books, and 4 must be selected; thus:

6C4 = 6!/(4! x 2!) = (6 x 5)/2! = 30/2 = 15 ways

Now we determine the total number of ways to select the books. There are 8 total books and 4 must be selected, thus:

8C4 = 8!/(4! x 4!) = (8 x 7 x 6 x 5)/4! = 7 x 2 x 5 = 70 ways

Thus, the number of ways to select at least one paperback book = 70 - 15 = 55 ways.

Answer: D