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One red bottle and five blue bottles are to be arranged in a straight

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Bunuel
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One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   22 Dec 2016, 02:21
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One red bottle and five blue bottles are to be arranged in a straight line from left to right on a windowsill. How many different unique arrangements are possible?

A. 6
B. 12
C. 30
D. 60
E. 120
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theLucifer
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   25 Dec 2016, 02:36
Possible positions :
RBBBBB
BRBBBB
BBRBBB
BBBRBB
BBBBRB
BBBBBR
Total no. of arrangements =6 position as above *5(each blue can be selected 5C1way)
=30


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rakaisraka
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   27 Dec 2016, 01:08
theLucifer wrote:
Possible positions :
RBBBBB
BRBBBB
BBRBBB
BBBRBB
BBBBRB
BBBBBR
Total no. of arrangements =6 position as above *5(each blue can be selected 5C1way)
=30


Sent from my Redmi Note 3 using GMAT Club Forum mobile app


Hi,
Shouldn't it be like
RBBBBB
6!/1!*5! = 6?
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KS15
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   09 Jan 2017, 06:20
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Bunuel wrote:
One red bottle and five blue bottles are to be arranged in a straight line from left to right on a windowsill. How many different unique arrangements are possible?

A. 6
B. 12
C. 30
D. 60
E. 120



Bunuel,

I think question should be 5 blue identical bottles-only in such cases can we get the answer as 6-otherwise if the bottles are not unique, then answer will be 6!=720
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JeffTargetTestPrep
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   11 Jan 2017, 08:07
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Bunuel wrote:
One red bottle and five blue bottles are to be arranged in a straight line from left to right on a windowsill. How many different unique arrangements are possible?

A. 6
B. 12
C. 30
D. 60
E. 120


This is a permutation with indistinguishable items problem. The order of the bottles matters, so it is a permutation problem; however, the 5 blue bottles are indistinguishable, so we will use the indistinguishable permutations formula. We need to determine in how many ways we can arrange one red bottle and five blue bottles in a straight line from left to right.

Since there are 5 indistinguishable blue bottles, we can arrange the bottles in 6!/5! = (6 x 5 x 4 x 3 x 2 x 1 )/(5 x 4 x 3 x 2 x 1) = 6 ways.

Answer: A
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KS15
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   11 Jan 2017, 09:05
JeffTargetTestPrep wrote:
Bunuel wrote:
One red bottle and five blue bottles are to be arranged in a straight line from left to right on a windowsill. How many different unique arrangements are possible?

A. 6
B. 12
C. 30
D. 60
E. 120


Jeff,

How did you say indistinguishable. Did you assume?
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JeffTargetTestPrep
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   11 Jan 2017, 10:58
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KS15 wrote:
JeffTargetTestPrep wrote:
Bunuel wrote:
One red bottle and five blue bottles are to be arranged in a straight line from left to right on a windowsill. How many different unique arrangements are possible?

A. 6
B. 12
C. 30
D. 60
E. 120


Jeff,

How did you say indistinguishable. Did you assume?


yes that is correct - I assumed all the red bottles were identical.
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KS15
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   11 Jan 2017, 11:04
Jeff,

I think we can't assume here. If all 6 bottles are not similar, then the answer will be different. The question needs to have the details.
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JeffTargetTestPrep
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   11 Jan 2017, 11:10
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KS15 wrote:
Jeff,

I think we can't assume here. If all 6 bottles are not similar, then the answer will be different. The question needs to have the details.


I agree - more information should be added.
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MathRevolution
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink] New post   22 Oct 2020, 04:50
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Red bottles: 1 and the Blue bottles: 5

Total bottles: 6

To arrange '6' bottles we have 6! ways.

As the bottles are same hence, \(\frac{6! }{ (1! * 5!)}\) = 6

Answer A
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Re: One red bottle and five blue bottles are to be arranged in a straight [#permalink]
22 Oct 2020, 04:50
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