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M10: Q24 - PS

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M10: Q24 - PS [#permalink] New post 22 Aug 2008, 14:41
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A box contains 10 tablets of medicine A and 15 tablets of medicine B. What is the least number of tablets that should be taken from the box to ensure that at least two tablets of each kind are among the extracted?

A. 12
B. 15
C. 17
D. 19
E. 21

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Re: M10: Q24 - PS [#permalink] New post 23 Aug 2008, 02:19
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The ans is C. In the worst case scenario, the first 15 tablets that you remove will be of medicine B. The next 2 tablets you remove will ensure that you have atleast 2 tablets of each kind among the extracted. So 15 + 2 = 17

A is not the answer beause you can remove 12 tablets from the box and all of them could be of medicine B.
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Re: M10: Q24 - PS [#permalink] New post 26 Sep 2009, 18:01
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What if I remove four tablets, N=4, and there happens to be two tablets of medicine A and two tablets of medicine B.

So the least number of tablets that ensure that at least two tablets of each kind are among the extracted tablets is four. Is this thinking correct? If so, I think the question should be reworded.

Someone please comment.
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Re: M10: Q24 - PS [#permalink] New post 29 May 2010, 20:50
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sid3699 wrote:
sid3699 wrote:
What if I remove four tablets, N=4, and there happens to be two tablets of medicine A and two tablets of medicine B.

So the least number of tablets that ensure that at least two tablets of each kind are among the extracted tablets is four. Is this thinking correct? If so, I think the question should be reworded.

Someone please comment.


Anyone :?:

Please? :|

Your approach above takes the best case scenario. However such an event happening (4 tablets = 2A + 2B) is extremely unlikely. The question asks for a guarantee - what is the least number of tablets you pull out so that you DEFINITELY GET 2 of each kind?
In this case you have to take the worst case scenario - lets say you keep pulling out tablets and you keep getting tablet B. In this case you can pull out 15 consecutive B's and then you will start getting A's.
So in the worst case you can pull out 15 straight B's and then the next 2 will be A. This satisfies the condition that you have 2 of each kind.
Hope it is clear now.
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Re: M10: Q24 - PS [#permalink] New post 10 Jun 2010, 10:14
To ensure two tablets from each of A and B are extracted:
15 of B's must have been exhausted.
Then followed by 2A's.
Correct response = 17 (C)
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Re: M10: Q24 - PS [#permalink] New post 14 Jun 2011, 22:58
I also got ans C, but my way of thinking was following-

we have 10 A's and 15 B's , so 150 variations and N!/2!(N-2)!
a)N!/2!(N-2)! =12!/2!*10!=66
b)105
c)136
d)171 (above 150, so exclude D and E)
since C is the greatest , then answ C

I agree that my way of thinking is more complicated than yours, but please let me know if it was a right one )) if not, where was I wrong?
p.s. I got my answ before looking at the right answ
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Re: M10: Q24 - PS [#permalink] New post 15 Jun 2011, 01:08
you look at the highest value:15

you can take 15 and still have them in the same color. so 15 only promises 2 of kind A.

now there are only kind B. taking 2 of kind B will ensuere: 2A and 2B (15+2=17)
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Re: M10: Q24 - PS [#permalink] New post 17 Jun 2011, 20:38
LalaB wrote:
I also got ans C, but my way of thinking was following-

we have 10 A's and 15 B's , so 150 variations and N!/2!(N-2)!
a)N!/2!(N-2)! =12!/2!*10!=66
b)105
c)136
d)171 (above 150, so exclude D and E)
since C is the greatest , then answ C

I agree that my way of thinking is more complicated than yours, but please let me know if it was a right one )) if not, where was I wrong?
p.s. I got my answ before looking at the right answ


How are you ensuring in this approach, that the pick has to be at least 2 each (4) from both A & B ? Shouldn't it be N!/ 4!(N-4)!
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Re: M10: Q24 - PS [#permalink] New post 18 Jun 2012, 05:33
Straight C. Taking 15 ensures that all B are taken (a number greater than 2 of course), then taking two additional tabs from the remaining, which consists of only A. Thus, 15 + 2= 17

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Re: M10: Q24 - PS [#permalink] New post 18 Jun 2013, 06:57
Expert's post
balboa wrote:
The box contains 10 tablets of medicine A and 15 tablets of medicine B. N tablets are taken from the box at random. What is the least possible N that ensures that at least two tablets of each kind are among the extracted?

(A) 12
(B) 15
(C) 17
(D) 19
(E) 21

[Reveal] Spoiler: OA
C

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I am trying to decide between a & c. Possible explanations?


BELOW IS REVISED VERSION OF THIS QUESTION:

A box contains 10 tablets of medicine A and 15 tablets of medicine B. What is the least number of tablets that should be taken from the box to ensure that at least two tablets of each kind are among the extracted?

A. 12
B. 15
C. 17
D. 19
E. 21

The worst case scenario will be if we remove all 15 tablets of medicine B first. The next 2 tablets we remove have to be of medicine A, so to guarantee that at least two tablets of each kind will be taken we should remove minimum of 15+2=17 tablets.

Answer: C.
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Re: M10: Q24 - PS   [#permalink] 18 Jun 2013, 06:57
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