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There are 8 books in a shelf that consist of 2 paperback [#permalink]
12 Nov 2007, 01:39

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Difficulty:

35% (medium)

Question Stats:

70% (02:18) correct
30% (01:47) wrong based on 210 sessions

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?

I was taking the approach of first finding out the number of ways of selecting 4 books out of 8, then finding the number of possibilities where no paperbacks were chosen, and subtracting the two.

If anyone can help with this approach to put down what the actual terms should be, thatd be very helpful to me.

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 ?

Soln: No. of combinations which have atleast one paperback is = Total no. of combinations - Combinations in which there is no paperback = 8C4 - 6C4 = 55 ways

Other way of doing this is finding out directly

No. of combinations which have atleast one paperback is = Total no. of combinations that have 2 paperback + total no. of combinations that have 1 paperback = 2C2 * 6C2 + 2C1 * 6C3 = 15 + 40 = 55 ways

Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
08 Feb 2012, 02:20

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Expert's post

pretttyune 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

Thanks a lot for your help in advance..

It's almost always better to solve "at least one" combinations/probability questions with {at least one}={total}-{none}.

Total # of selections of 4 books out of 8: C^4_8=70; Total # of selections of 4 hardback books out of 8 (so none paperback): C^4_6=15;

Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
08 Feb 2012, 09:00

# of selections of books with no condition = 8C4 = 70 # of selections of books with no paperback book = 6C4 = 15 # of selections of books with at least one paperback book = 70 -15 = 55

Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
25 Jan 2013, 07:20

Hmm....so when I did this I made an error that might have carried over from using combinatorics from probability. I found the total as 8C4 and to find the amount to subtract from it I had (6C4)(2C0). Why don't we multiply by the books not chosen and what's would it mean if we did?

I'm trying to understand a fundamental flaw I made here _________________

Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
25 Jan 2013, 07:33

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Expert's post

manimgoindowndown wrote:

Hmm....so when I did this I made an error that might have carried over from using combinatorics from probability. I found the total as 8C4 and to find the amount to subtract from it I had (6C4)(2C0). Why don't we multiply by the books not chosen and what's would it mean if we did?

I'm trying to understand a fundamental flaw I made here

You'd get the same result because 2C0=2!/(2!*0!)=2/(2*1)=1: there is one way to choose 0 books from 2 books.

Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
02 Feb 2014, 18:08

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