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There are 9 books on a shelf, 7 hard cover and 2 soft cover.

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There are 9 books on a shelf, 7 hard cover and 2 soft cover. [#permalink]

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02 Oct 2003, 08:48
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There are 9 books on a shelf, 7 hard cover and 2 soft cover. How many different combinations exist in which you choose 4 books from the 9 and have at least one of them be a soft cover book?
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shubhangi

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02 Oct 2003, 09:13
Can u explain in detail.....i didnt get it amarsesh
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shubhangi

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02 Oct 2003, 09:21
shubhangi wrote:
Can u explain in detail.....i didnt get it amarsesh

4 books can be chosen from 9 books in = C(9,4)

There will be no soft cover books if all the 4 books chosen are hard covers and they can be chosen in C(7,4)

So, there will be atleast one soft cover in C(9,4) - C(7,4) combinations

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02 Oct 2003, 09:30
agree with amarsesh.
shubh if it is still not clear let us know (note: be more specific as to which part of problem u are not getting).
thanks

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02 Oct 2003, 09:31
Okay! got it
But what will be the answer??
Thanks
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shubhangi

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02 Oct 2003, 09:33
Hey vicks!!
Actually , i am not getting the calculation part.I mean what will be the answer..becoz in book its 91
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shubhangi

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02 Oct 2003, 09:35
we just need to calculate it:
9C4 = 9*8*7*6/1*2*3*4 = 126 -----------a
7C4 = 7C3 = 7*6*5/1*2*3 = 35 ----------b
right...

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02 Oct 2003, 09:37
shubhangi wrote:
Hey vicks!!
Actually , i am not getting the calculation part.I mean what will be the answer..becoz in book its 91

C(n,k) = n! / (k! * (n-k)!)
where n! = n * (n-1)! for n > 1
= 1 for n=0

C(9,4) = 9!/(4! * 5!) = 126
C(7,4) = 9!/(4! * 3!) = 35

So, answer = 126-35 = 91

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02 Oct 2003, 09:48
C(9,4) = 9!/(4! * 5!) = 126
So, that means...this will look like this:
9*8*7*6*5/4*3*2*5*4*3*2*1...right??
but calculate it.......its not 91
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shubhangi

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02 Oct 2003, 10:00
I think i am no good in permutations and combination
can u suggest some thing to improve it..
thanks
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shubhangi

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02 Oct 2003, 10:07
shubhangi wrote:
I think i am no good in permutations and combination
can u suggest some thing to improve it..
thanks

C(9,4) = 9!/(4! * 5!) = (9*8*7*6*5*4*3*2*1)/(4*3*2*1)*(5*4*3*2*1) = 126
C(7,4) = 7!/(4! * 3!) = (7*6*5*4*3*2*1)/(4*3*2*1)*(3*2*1) = 35

Hope that clears things for you.

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02 Oct 2003, 10:11
yes!! ofcourse
i should really practise per/com..
thanks a lot!!
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shubhangi

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02 Oct 2003, 10:14
why don't u try the permutation/combinations stuff on this website.
U can also consult any basic book on combinations/permutations or search on google for some stuff. I guess someone posted in share GMAT expe. section a link to some basic permutation/combo site. Check it out
Remember, u would want to be thorough in this area cause more and more questions based on them have begun to appear on GMAT
also 9! = 1*2*3...*8*9
it's pretty late here... time to sleep...
nowadays even in sleep i get these tormenting SC's...
thanks

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02 Oct 2003, 10:17
Hey !! vicky
Thanks a lot
i will try out.....as u said..
Tell me when can we share more problmes
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shubhangi

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02 Oct 2003, 10:17
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