# Thread: Permutations / Combinations Question

1. ## 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.

2. my guess would be...

$\binom{2}{1}\times\binom{6}{3}+\binom{2}{2}\times\ binom{6}{2} = 55$

3. 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?
$\sum\limits_{k = 1}^2 {\binom{2}{k}\binom{6}{4-k}}$

4. 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.

${8\choose4}=70$

${6\choose4}=15$

70-15=55

Another way:

If we have 2 hardcovers and 2 paperbacks:

${6\choose2}$

If we have 3 hardcovers and 1 paperback:

$2\cdot{6\choose3}$

Add them and get 55

5. Thanks everyone! Very helpful. I had one more question, kind of similar, but I don't know how to solve this one mathematically without drawing it out ...

Six three-representative delegations attend an international conference. Teh reps shake hands when they are introduced to one another, how many handshakes are possible if each delegate shakes hands only once with every other attendant except with those of his/her delegation

As always, any help is greatly appreciated.

6. Nevermind. Figured it out. :-)