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 6 (so none paperback): \(C^4_6=15\);

{at least one} = {total} - {none} = 70 - 15 = 55.

Answer: D.
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i would go about it this way:

total - all hardback
total = 8C4
hardback = 6C4

8C4 - 6C4
70-15 = 55

No of combinations = p(1) + p(2)

= (2C1 * 6C3) + (2C2 * 6C2)

= (2*6*5*4)/(3*2*1) + (6*5)/2

= 40 + 15

= 55

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

The selection should at least contain one PAPERBACK book. There are a total of 2 PB and 6 HB books are available.

The combinations that at least one PB book will come out are: PHHH & PPHH

1. PHHH = 2C1 * 6C3 = 2 * (5*4) = 40
2. PPHH = 2C2 * 6C2 = 1 * ((6*5) / 2) = 15

In total 40+15 = 55 Ways

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.

Hope it's clear.
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There are two scenarios that fit the question: either one paperback or two is selected.

If one is selected, then the total number of combinations is 2C1*6C3 + 2C2*6C2

2C1*6C3:
2*6!/(3!*3!) = 2*6*5*4/3*2 = 40

2C2*6C2
1*6*5*4*3*2/(2*4*3*2) = 15

40+15 =55

Could someone provide a group of similar practice problems? Would really appreciate the help on this. Thanks

21. Combinatorics/Counting Methods

• Theory
  Math: Combinatorics
  Combinatorics Made Easy!
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  Combinations while dealing with similar and dissimilar things
  When Permutations & Combinations and Data Sufficiency Come Together
  
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    Advanced Topic: Probability and Combinations Questions With Solutions
    Letter Arrangements in Words
    Seating Arrangements in a Row and around a Table
    Constructing Numbers, Codes and Passwords
    DS Questions
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22. Probability

• Theory
  Math: Probability
  Probability Made Easy!
  GMAT Tip of the Week: 99 Problems But Probability Ain't One
  
  • Questions
    Probability and Geometry Problems
    Advanced Topic: Probability and Combinations Questions With Solutions
    DS Questions
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For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
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We can use the equation:

The number of ways in which at least 1 paperback book is selected = The total number of ways to select 4 books – The number of ways in which no paperback books are selected

The number of ways in which no paperback books are selected is equivalent to the number of ways in which all 4 books selected are hardcover. Let’s determine that now. There are 6 hardback books, and 4 must be selected; thus:

6C4 = 6!/(4! x 2!) = (6 x 5)/2! = 30/2 = 15 ways

Now we determine the total number of ways to select the books. There are 8 total books and 4 must be selected, thus:

8C4 = 8!/(4! x 4!) = (8 x 7 x 6 x 5)/4! = 7 x 2 x 5 = 70 ways

Thus, the number of ways to select at least one paperback book = 70 – 15 = 55.

Answer: D
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Hi Bunuel Bunuel,
Here why cant we do like below -
Total = 8*7*6*5
Only Hardback = 6*5*4*3

Required = Total - Hardback

Say we have A, B, C, D, E, F, G, and H books.

With 8*7*6*5 you'd get duplications in 4-book selections. If we choose A first, then choose B, then choose C and finally choose D, you'd get the same four books as if we choose D first, then choose A, then choose B, and then choose C. Note that the order of the books does not matter. Your method assumes that the order of the books does matter - you're choosing one book as the "first book", another as the "second book" and so on. Thus you get duplications. In order to get rid of them, you should divide 8*7*6*5 by 4!.
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In general, when it comes to counting questions involving "at least," we can use the following property:
Number of outcomes that satisfy the restriction = (number of outcomes that ignore the restriction) - (number of outcomes that BREAK the restriction)

So we get:
Number of combinations with at least one paperback = (total number of 4-book combinations) - (number of 4-book combinations that DON'T have at least one paperback)

In other words:
Number of combinations with at least one paperback = (total number of 4-book combinations) - (number of 4-book combinations with ZERO paperbacks)

total number of 4-book combinations
There are 8 altogether, and we must choose 4 of them
Since the order in which we select the books does not matter, we can use combinations.
We can select 4 books from 8 books in 8C4 ways (= 70)

number of 4-book combinations with ZERO paperbacks
In order to get ZERO paperbacks, we'll select 4 books from the 6 hardback books only.
We can select 4 books from 6 books in 6C4 ways (= 15)

So, the number of combinations with at least one paperback = (70) - (15) = 55

Answer: D
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