# Permutations / Combinations Question

• Jul 17th 2009, 03:09 PM
dkn2101
Permutations / Combinations Question
The following problem I just can't seem to solve:

From a set of 6 hardcover and 2 paperback books, how many different 4-book sets can be formed if each 4-book set must contain at least one paperback?

The answer says 65 ... but I have no idea how they're getting to this answer. I tried multiplying (6 3)*(2 1) (6 choose 3, 2 choose 1)... but that can't be right because the 4 book set must contain AT LEAST one paperback, so it can also contain 2.

Any help here would be much appreciated. Thanks.
• Jul 17th 2009, 03:38 PM
pickslides
my guess would be...

$\displaystyle \binom{2}{1}\times\binom{6}{3}+\binom{2}{2}\times\ binom{6}{2} = 55$
• Jul 17th 2009, 03:40 PM
Plato
Quote:

Originally Posted by dkn2101
From a set of 6 hardcover and 2 paperback books, how many different 4-book sets can be formed if each 4-book set must contain at least one paperback?

$\displaystyle \sum\limits_{k = 1}^2 {\binom{2}{k}\binom{6}{4-k}}$
• Jul 17th 2009, 03:47 PM
galactus
Maybe I am counting wrong, but I get 55.

The thing to do is count the total number of ways to choose 4 from 8.

Then, find the number ways of choosing a 4 book set with NO paperback and subtract them.

$\displaystyle {8\choose4}=70$

$\displaystyle {6\choose4}=15$

70-15=55

Another way:

If we have 2 hardcovers and 2 paperbacks:

$\displaystyle {6\choose2}$

If we have 3 hardcovers and 1 paperback:

$\displaystyle 2\cdot{6\choose3}$