dave13 wrote:
here is corrected version of my solution
many thanks !
by the way is there a faster method ?
\(4C2*6C3 = 6*20 = 120\) (# OF WAYS TO CHOOSE 2 PAPERBACKS FROM 4 AND 3 HARDBACKS FROM 6)
\(4C3*6C2 = 4*15 = 60\) (# OF WAYS TO CHOOSE 3 PAPERBACKS FROM 4 AND 2 HARDBACKS FROM 6)
\(4C4*6C1 = 1*6 =6\) (# OF WAYS TO CHOOSE 4 PAPERBACKS FROM 4 AND 1 HARDBACK FROM 6)
\(4C1*6C4 = 4*15 =60\) (# OF WAYS TO CHOOSE 1 PAPERBACK FROM 4 AND 4 HARDBACKS FROM 6)
# WAYS TO SELECT 5 BOOKS CONTAINING AT LEAST ONE PAPERBACK AND ONE HARDBACKS = \(120+60+60+6 = 246\)
Hey
dave13The easiest way to do this problem is using the method
chetan2u has solved
this question in. Am merely building on his solution.
There are 4 paperback books and 6 hardback books of which 5 books are to be selected
There are \(C_5^{10} = 252\) ways of selecting these books.
Now we need to reduce the options where there are only paperbacks/hardbacks. Only
paperbacks are not possible because there are only 4 paperbacks. You will have to have
one hardback at the minimum to select the 5 books.
But there are 6 ways in which the hardbacks can be selected.
For our convenience, let the hardback books be H1, H2, H3, H4, H5, H6
We need 5 books, possible options of which are 6 which are listed below
H1, H2, H3, H4, H5
H1, H2, H3, H4, H6
H2, H3, H4, H5, H6
H1, H3, H4, H5, H6
H1, H2, H3, H5, H6
H1, H2, H4, H5, H6
Therefore, there are a total of 252 - 6 =
246 ways(Option D) in which the books can be chosen
Hope this helps you!
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