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how i should approach this one

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how i should approach this one [#permalink] New post 19 Nov 2016, 09:31
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hi,

how many no.of ways 6 rings can be worn in 4 fingers?

its a combination question. but i dont know how to techniques work this out?

thank you
siva
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Re: how i should approach this one [#permalink] New post 19 Nov 2016, 10:37
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ssiva wrote:
hi,

how many no.of ways 6 rings can be worn in 4 fingers?

its a combination question. but i dont know how to techniques work this out?

thank you
siva


Hey this is a bad question because we do not know if the rings are distinct and fingers are distinct.

For example if all the rings are same but fingers are not then the problem is to arrange 6 things in 4 places.

Let the rings in each fingers be x1, x2 .... x4.

So essentially you are asking how many solution to the equation

\(x1 + x2 + x3 + x4 =6\)

The number of different ways to distribute 6 indistinguishable rings into 4 distinguishable fingers is C(6+4-1,4-1).

or C(9,3)=\(\frac{9!}{3!6!}\)=84
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Re: how i should approach this one [#permalink] New post 20 Nov 2016, 16:07
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ssiva wrote:
hi,

how many no.of ways 6 rings can be worn in 4 fingers?

its a combination question. but i dont know how to techniques work this out?

thank you
siva


I agree - this is too ambiguous to be a GRE-worthy question.
What is the source of this question?

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
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Re: how i should approach this one [#permalink] New post 18 Dec 2016, 15:43
I think the answer is 360. You have six rings and four choices of where to put them. On the first finger you have six choices for the ring so that is 6, 5 choices for the second finger, 4 ring choices for the third finger and 3 choices for the fourth finger. Imagine it like a tree with branches of different possibilities spreading out from the tree, depending on which option you pick each time. This means you have to multiply the possibilities rather than adding them. 6*5*4*3 is 360.
Re: how i should approach this one   [#permalink] 18 Dec 2016, 15:43
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