There are 8 books on a shelf, of which 2 are paperbacks and 6 are hardbacks. Pick 4 books.?

Question

There are 8 books on a shelf, of which 2 are paperbacks and 6 are hardbacks. How many possible selections of 4 books from this self include at least one paperback?

Please provide answer and your explanation. THANX

Answer 1

OK this is long winded, but name the books

P1,P2 and H1,H2,H3,H4,H5,H6  (paperback and hardback)

Start with  P1. that's your one paper back so you need 3 hardback, in various combinations.

1,2,3  1,2,4  1,2,5  1,2,6, 1,3,4  1,3,5  1,3,6  1,4,5 1,4,6  1,5,6

2,3,4   2,3,5  2,3,6  2,4,5  2,4,6

3,4,5   3,4,6

4,5,6

Notice combinations such as 1,3,2 are misse dout becase it is a repeat of 1,2,3 and you haven't stated that order of books is relevant, so you count that as a single combination and not as seperate ones.

Therefore with paperback 1, you have 18 cominations. with paper back 2 you will also get 18 as the hardback numbers remain unchanged and still have the same amount of combinations

using only one paper back you get 36 cominations,

however as it says at least 1 paperback, so therefore you can include 2 paperbacks so you have to include combinations for 2 paperbacks also

which means you only need to choose 2 hardbacks

1,2  1,3  1,4  1,5  1,6

2,3  2,4  2,5  2,6

3,4  3,5  3,6

4,5  4,6

5,6

15 combinations including 2 paperback

total of 51 combinations altogether.

Hope it helps =]

btw i'm not a geek, we just used to have these questions given to us often aswell.. i feel your pain ¬¬

Answer 2

I used the formula of combination. Combination is the number of ways of selecting r objects from n without regard to order. The number of combinations of n distinct objects taken r at a time is nCr = n! /[r!(n-r)!]

First, I used the formula to determine how many possible combinations there are. there are 8 distinct objects taken 4 at a time which means that n=8 and r=4. Using the formula (it's tiring to type out), you can get the answer 70.

Now, there are only three possibilities. Since there are only 2 paperbacks, the three possibilities of combinations are as follows.

a. 2 Paperbacks and 2 Hardbacks

b. 1 Paperback and 3 Hardbacks

c. 0 Paperback and 4 Hardbacks

There can't be an option d because it's impossible to have 3 Paperbacks and 1 Hardback since there are only 2 Paperbacks.

What I did next was to compute the number of possibilities each option could occur.

For a. 2 Paperbacks and 2 Hardbacks, I used the formula of combination again where n = 6 and r = 2 because I wanted to know how many combinations of the 2 Hardbacks can be made (since there are 6 Hardbacks. Using the formula, I acquired the answer 15.

For b. 1 Paperback and 3 Hardbacks, I used the formula of combination where n = 6 (Hardback books) taken 3 (r=3) at a time. Using the formula, I acquired the answer 20 but since there are 2 paperbacks and they are distinct, I multiplied the combinations by 2 thereby getting a final answer of 40. (because it can be H1, H2, H3, P1 and H1, H2, H3, P2. If I didn't multiply the original answer by 2, I am assuming that P1 and P2 are the same).

Therefore, adding the possibilities, the answer is 55 (40+15).