Carcass wrote:
There are 8 books in a shelf that consist 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
In general, when it comes to counting questions involving "at least," we can use the following property:
Number of outcomes that satisfy the restriction = (number of outcomes that ignore the restriction) - (number of outcomes that BREAK the restriction)So we get:
Number of combinations with at least one paperback = (total number of 4-book combinations) - (number of 4-book combinations that DON'T have at least one paperback)In other words:
Number of combinations with at least one paperback = (total number of 4-book combinations) - (number of 4-book combinations with ZERO paperbacks)total number of 4-book combinationsThere are 8 altogether, and we must choose 4 of them
Since the order in which we select the books does not matter, we can use combinations.
We can select 4 books from 8 books in 8C4 ways (=
70)
number of 4-book combinations with ZERO paperbacksIn order to get ZERO paperbacks, we'll select 4 books from the 6 hardback books only.
We can select 4 books from 6 books in 6C4 ways (=
15)
So, the
number of combinations with at least one paperback = (70) - (15) =
55Answer: D
25 Feb 2022, 07:51 
75% (01:46) correct
