What is the answer to this question, and how is it solved?

You're given 10 books, 6 paperback and 4 hardback.  How many
combinations can you make
with at least 1 hardback?

I calculate it to be 960.
My logic is as follows.
6 paperbacks. Possible combinations are 1(of 6 books), 6 (of 5 books),
15, 20, 15,6 0f 4, 3,2,1 books respectively (using formula nCk =
n!/k!(n-k!). This total comes to 63.
4 hardbacks. Possible combos are 1+4+6+4 for 4,3,2,and 1 book
respectively. A total of 15.
Multiply 63 by 15; add 15 for combos including only hardbacks.

I calculate it to be 784.

You can have either 1, 2, 3, or 4 hardbacks and anywhere from 0 to 6 paperbacks.
Below is a table with the number of combinations for each.

							
No. of	Number of Paperbacks					Total
Hbacks	0	1	2	3	4	5	6	Combinations
	
	Number of Combinations						
1	1	2	3	4	5	6	7	28
2	1	3	6	10	15	21	28	84
3	1	4	10	20	35	56	84	210
4	1	5	15	35	70	126	210	462
								
	4	14	34	69	125	209	329	784

As an example the number of combinations for 4 hardback and 6 paperbacks is

10!/[(4!)*(6!)] = 210


Of course, if the order in the combination didn't matter, there would
be only 28 combinations.  Anywhere from 1 to 4 hardbacks combined with
anywhere from 0 to 6 paperbacks:  4 x 7 = 28.  I assumed above, that
order does matter.

If it still matters to anyone, I'd like to point out that the first
response (960) is correct. I can elaborate on why this is so - just
let me know by adding another comment.

Have a good day.

Ok, I'll bite.  Why?

ansel001 - 

   Hello. Let's see if we can't get to the bottom of this and figure
out whether 960 is indeed the correct answer.

   Just a few notes before I start:
   1) I don't really understand how you arrived at your numbers so I
won't approach the question from that angle - I'll just try to flesh
out the original answer which is rather succinct.
   2) This explanation might take a few posts on my part.
   
   Permutation refers to all the possible arrangements of a group. So,
if you have the elements abc you would have 6 permutations:
abc/acb/bac/bca/cab/cba. In permutation, order does matter - different
orders of the same elements are counted separately.
   Combination refers to just to the unique elements - abc is one
combination regardless of arrangement whereas abd would be a different
combination. Order here does not matter.
  
   In this case we are looking for combinations - which makes sense
because, in a real world situation, you wouldn't say that Jack and
Jill have different books because Jack has Hardback X and Paperback Y
while Jill has Paperback Y and Hardback X.

   More tomorrow...

(I will be using a, b, c, and d to represent the hardbacks and p, q,
r, s, t and u to represent the paperbacks)

   Ok, so each and every combination that answers the conditions of
the problem will consist of some combination of hardbacks and some
combination of paperbacks - please note, that I am, for the moment,
ignoring the combinations that can be made without using any
paperbacks (hardbacks alone).

   You can kind of mentally divide the final combination into two
parts - a hardback combination component and a paperback component.
   For instance, a/b coupled with p/q would make a valid final
combination as would a/b coupled with p/r as would a/b coupled with
p/q/r as would a/b/c coupled with p/q, etc.

   In other words, we need to determine all the combinations of the
hardbacks, all the combinations of the paperbacks and then all the
combinations of both of these.

   The hardback combinations are:

   for one hardback - a/b/c/d (4 combinations)
   for two hardback books - ab/ac/ad/bc/bd/cd (6 combinations) 
   for three hardback books - abc/abd/acd/bcd (4 combinations)
   for four hardback books - abcd (1 combination)

   The paperback combinations are:

   for one paperback - p/q/r/s/t/u (6 combinations)
   for two paperbacks - pq/pr/ps/pt/pu/qr/qs/qt/qu/rs/rt/ru/st/su/tu
(15 combinations)
   for three paperbacks -
pqr/pqs/pqt/pqu/prs/prt/pru/pst/psu/ptu/qrs/qrt/qru/qst/qsu/qtu/rst/rsu/rtu/stu
   (20 combinations)
   for four paperbacks -
pqrs/pqrt/pqru/pqst/pqsu/pqtu/prst/prsu/prtu/pstu/qrst/qrsu/qrtu/qstu/rstu
   (15 combinations)
   for 5 paperbacks - pqrst/pqrsu/pqrtu/pqstu/prstu/qrstu (6 combinations)
   for 6 paperbacks - pqrstu (1 combination)
   
   I typed out the combinations just so they're there, but you don't
really need to write them out - they can be determined mathematically.

Thither,

Now I see what has happened.  We have each interpreted what
"combination" means differently, and answered the question as we
interpreted it, correctly.

You assumed that each hardback and paperback book is differentiated
and it matters which hardback or paperback you are referring to.  But
once the individual hardbacks or paperbacks are selected, the
particular arrangement of the books didn't matter.

I interpreted the question along the lines in which you often see it. 
Suppose you have 4 red balls and 6 blue balls.  How many combinations
can you make that include at least one red ball.  In this case, the
red balls are undifferentiated as are the blue balls.  You don't care
which red balls or blue balls you are talking about, only how many of
each color you have.  And the arrangement matters.

For example, with 2 red balls and 2 blue balls you would have 6 combinations.

1.  RRBB
2.  RBRB
3.  RBBR
4.  BRRB
5.  BRBR
6.  BBRR

I interpreted hardbacks and paperbacks in the same sense as the red
and blue balls.  It only matters how many hardbacks and paperbacks you
have, not which ones.  And arrangement matters.

I answered my question correctly and you answered your question
correctly.  The question should have been stated more clearly.

Cheers.

Thither,

It occurs to me this morning that using you definition of
"combination" there is an easier way to solve this.

That would be:

The total number of combinations of all 10 books
Minus the number of combinations of paperbacks only

The number of combinations of n books is 2^n-1

So the above formula yields:

(2^10-1) - (2^6-1) = 1,023 - 63 = 960

ansel001 -

   I'll have to consider your last solution (it's been a while since
statistics class).
   I just wanted to mention that yes, we were answering different
questions in the original answers.
   The question should have been more explicit - although it is my
understanding that when a question references things that are easily
differentiable (like books with their titles) that it should be
thought of differently than questions which involve colored balls,
socks, etc. (which are essentially the same).
   Of course, I'm not sure exactly how much of a convention this is.

   Have a good day.

Dear ansel001,
I approached the problem based on the meaning of the word combination
in math (as distinct from permutation).
Thither, I liked the way you explained the answer. I guess my post was
comprehensible to someone who already knew how to do it, and "succint"
(as you kindly put it!!) aka mysterious to someone who didn't!
I am intrigued ansel, by the 10 feb solution. I cant get my head
around it, the numbers do work, but I get a nagging feeling it may be
coincidental (like 2+2 = 2x2)

Person asks a question on Google Answers.
Person receives answer (for free no less).
Person confidently moves onward towards ever-greater accomplishments.

And here we are...

ansel's last solution is indeed valid.

The number of unique combinations that come from n elements is 2^n-1
(as I understand it).

So, in carryou's original answer, there's no need to add up to find
the combinations. You can simply say:

(combinations using only hardbacks) + (all combinations of
hardbacks)*(all combinations of paperbacks)

or 

2^4-1 + (2^4-1)(2^6-1) which equals 2^10-2^6 = 960.

Ansel's solution approaches it from the other direction:

Total combinations of the 10 books would be 2^10-1 (of course this
includes combinations which don't include a hardback).

But this total is simply:

(combinations using only hardbacks) + (all combinations of
hardbacks)*(all combinations of paperbacks) + (combinations using only
paperbacks)

or

2^4-1 + (2^4-1)(2^6-1) + (2^6-1) = 2^10-1

So, 2^10-1 - 2^6-1 = 2^4-1 + (2^4-1)(2^6-1)

Using letters...

Original solution H+(H)(B)= answer

ansel's solution H+(H)*(B)+B = 2^10-1 so 2^10-1 - (B) = H+(H)*(B) = answer

Everything's copacetic, right?

By the way, I have awarded both of you a gold star for your efforts.

Thither
I had to look up copacetic. Shall now wait for a proper place to use it.
BTW, 2 gold stars to you!
Carryou