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

Solution:

Total possible combinations of books = C(8,4)

Total possible combinations without paperback books C(6,4)

Total possible combinations with at least 1 paperback book: C(8,4)-C(6,4)=55

Another approach:

Combinations with the first book:
6C3 = 20

Combinations with the second book:
6C3 = 20

Combinations with both books:
6C2 = 15

Total = 20+20+15 = 55

Regards

PHHH + PPHH

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

= 2 * 6!/3!3! + 1 * 6!/4!2!

= 2* 6*5*4/3! + 6 * 5/2

= 40 + 15 = 55
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I came up with a different (more complicated?) approach.

FIRST STEP:
How many possible selections of 4 books from this self including NO paperback?
6/8 * 5/7 * 4/6 * 3/5 = 3/14

SECOND STEP:
Subtract 3/14 from 1 (or 14/14) => 14/14 - 3/14 = 11/14

THIRD STEP:
The only answer choice with 11 as a factor is 55.

Ans. C

Do i miss the answer choices?

total combinations: \(C^4_8 = 70\)
No Paperback: \(C^4_6 = 15\)

Total - No Paperback = At least 1 Paperback
70 - 15 = 55 combinations

# 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

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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# of ways of choosing 4 books from 8 total books = 8C4 = 70
# of ways of only choosing hardback books = 6C4 = 15

Therefore, # of ways of choosing 4 books with at least 1 paperback book = 8C4 - 6C4 = 55 (D)

When I first read the question I understood that we need to find the number of ways to select 4 books with at least 1 of them being a paperback, instead of calculating the number of possible selections, were PHHH = HPHH

Would the answer to this other question be correct, as this might show up on the exam?
2 Paperbacks: 2*1*6*5 = 60
1 Paperback : 1*6*5*4 = 120
= 180

First thing to notice when you look at this question: we're dealing with permuations/combinations.
Which to use? Ranking/order doesn't matter here - which means we use combinations.

Next thing to notice: at least one - i don't know anyone who likes these questions!
Strategy should be TOTAL - NONE = AT LEAST 1 (our desired solution)

TOTAL = 8c4 (8 books 4 being selected) = 70
NONE = 6c4 (Only 6 books because we aren't including paperback, 4 being selected) = 15
70-15 = 55

CHOOSE D

ON TO THE NEXT

a method to solve this problem
so there are 2 possibilities
either : 1 PB and 3 HB = 2*6*5*4/3! = 40
OR : 2 PB and 2 HB = 2*1*6*5/2!*2! = 15

so we have to add both the possibilities to get the final answer

40+15=55

Hi Bunuel,

Using PB: Paperbacks ; HB : Hardbacks

If the question were - "How many possible selections of 4 books include both the PB" then is why is my following solution wrong ?

Total possibility .: C(8,4)
Possibility of 2 PB = Total possibility - possibility of no PB => C(8,4)- C(6,4) [C(6,4) -> removing 2 PB from the lot will leave only 6 books. So selecting 4 books from 6].

So answer is C(8,4)- C(6,4) = 70-15 = 55

1. Method :- 8c4 - 6c4 = (Total - All four book selected from 6 hard copy) - This will give atleast one paper back selected.

2.Method :- 6c2*2c2 + 6c3*2c1 = (PPHH) + (PHHH)