A bag contains 10 balls, each of which is labeled with a : PS Archive
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# A bag contains 10 balls, each of which is labeled with a

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Senior Manager
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A bag contains 10 balls, each of which is labeled with a [#permalink]

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03 Jan 2008, 08:10
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A bag contains 10 balls, each of which is labeled with a distinct
prime number. If 2 balls are to be removed from the bag and
the numbers on the balls are multiplied together, how many
different products are possible?

(A) 10
(B) 45
(C) 50
(D) 90
(E) It cannot be determined from the infomation given.
Director
Joined: 09 Jul 2005
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03 Jan 2008, 09:00
All the possible combinations of ten elements combined in groups of two elements which is 10!/[8!2!]=45 B
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03 Jan 2008, 09:24
Yup, same thing as thinking "I have 10 choices for my first pick and 9 for my second"

10*9 = 90

but the order doesn't matter since they're just being multiplied by one another. So divide by 2 to get rid of any duplicates (e.g. (3,5) and (5,3))

90/2 = 45

The original combination approach is the more correct way to go about it, but using permutations and modifying it works just as well.
SVP
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03 Jan 2008, 17:05
I went for 90

forgot that the question asked for "different" products ! The answer should be half of 90 . i.e. 45
CEO
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04 Jan 2008, 00:06
10!/8!2! --> 45

You can check this by lining up the primes: well just call them A-J

ABCDEFGHIJ
ABCDEFGHIJ draw lines from A to B to C etc.. then do the same for B but not drawing a line to either B or A.
Manager
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05 Jan 2008, 06:09
45,
there are two places available and we know= possible first place x possible second place
so, 10 x 9, divide by 2 and get 45
Re: PS Question   [#permalink] 05 Jan 2008, 06:09
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