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M26-13

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M26-13  [#permalink]

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New post 16 Sep 2014, 01:25
5
28
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A
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D
E

Difficulty:

  95% (hard)

Question Stats:

32% (02:07) correct 68% (02:18) wrong based on 136 sessions

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Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?

A. 30
B. 60
C. 120
D. 240
E. 480

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New post 16 Sep 2014, 01:25
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Official Solution:

Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?

A. 30
B. 60
C. 120
D. 240
E. 480


Seems tough and complicated but if we read the stem carefully we find that the only way both conditions can be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in \(\frac{5!}{2!*2!}=30\) ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R), then total # of arrangements is \(30*2=60\).


Answer: B
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Re: M26-13  [#permalink]

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New post 02 Dec 2018, 21:23
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Quote:
I got the use of formula. However, why we multiply by 2 at the end, is the part I am NOT getting (never understood such thing till date).

Request your help Bunuel.


Because there are 2 ways to arrange 5 red marbles (R*R*R*R*R* or *R*R*R*R*R)
Each way we have 30 ways to arrange 5 remaining marbles including 2 blue, 2 green and 1 yellow :D

Hope it helps!
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M26-13  [#permalink]

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New post 09 Dec 2018, 02:30
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AashiS wrote:
Why cant we use 10! as there are total 10 slots and all marbles will repeat which can come in the denominator.
5!*2!*2!

Do you mean your formula is \(\frac{10!}{5!*2!*2!}\)?

Let me try to answer this. You can't do that because you have to "arrange all marbles in a row that no two adjacent marbles are of same color and the first and the last marbles are of different colors" and using the formula above, you can't satisfy these conditions.

There are two ways to arrange 5 red marbles (R*R*R*R*R* or *R*R*R*R*R) but using above formula, it actually counts a lot more cases such as RRRRR*****, RR**RR***R, etc which is not suitable to this question => so you have to arrange red marbles first because the number of red marbles is more than the number of two remaining color.

Hope it's clear.
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Re: M26-13  [#permalink]

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New post 06 Sep 2016, 10:03
Hi Bunuel,

Can you elaborate on the 5!/2!*2! portion of the answer? If there are three color marbles, how do we know how to set this up? Thanks in advance.
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Re: M26-13  [#permalink]

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New post 27 Dec 2016, 03:40
Hi Buenel,

Can you please post more sums like this one for us to practise?

Many Thanks
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Re: M26-13  [#permalink]

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New post 07 Jun 2018, 21:24
Bunuel wrote:
Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?

A. 30
B. 60
C. 120
D. 240
E. 480


It is not mentioned that The marbles are identical. They may have the same color but it may be possible that they have different numbers or something different that makes each arrangement unique, so in that case the solution changes.
My doubt is if it is not explicitly mentioned that the marbles are identical then should we assume they are? (i mean looking at the answer options it suggest that balls are identical)
I know its a not a good query but Kindly suggest a solution
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New post 03 Aug 2018, 10:51
dpgiii wrote:
Hi Bunuel,

Can you elaborate on the 5!/2!*2! portion of the answer? If there are three color marbles, how do we know how to set this up? Thanks in advance.


5!/2!*2! is because there are two blue and two green marbles.

5! = # of ways to set up five marbles
2! = repetition of two blue marbles
2! = repetition of green marbles.

You divide by # of same kinds in factorial in the denominator when counting the total number of ways.
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Re: M26-13  [#permalink]

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New post 24 Sep 2018, 01:26
I got the use of formula. However, why we multiply by 2 at the end, is the part I am NOT getting (never understood such thing till date).

Request your help Bunuel.

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M26-13  [#permalink]

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New post 08 Dec 2018, 23:21
Why cant we use 10! as there are total 10 slots and all marbles will repeat which can come in the denominator.
5!*2!*2!
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New post 09 Dec 2018, 03:39
Samine wrote:
AashiS wrote:
Why cant we use 10! as there are total 10 slots and all marbles will repeat which can come in the denominator.
5!*2!*2!

Do you mean your formula is \(\frac{10!}{5!*2!*2!}\)?

Let me try to answer this. You can't do that because you have to "arrange all marbles in a row that no two adjacent marbles are of same color and the first and the last marbles are of different colors" and using the formula above, you can't satisfy these conditions.

There are two ways to arrange 5 red marbles (R*R*R*R*R* or *R*R*R*R*R) but using above formula, it actually counts a lot more cases such as RRRRR*****, RR**RR***R, etc which is not suitable to this question => so you have to arrange red marbles first because the number of red marbles is more than the number of two remaining color.

Hope it's clear.


Okay Got it know, Red are most in number and have to be arranged next to different color marble, after red marbles can arrange rest..

Thanks So Much!
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Re: M26-13  [#permalink]

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New post 01 Mar 2019, 01:38
When we divide 5! by 2!*2!, what does the denominator imply? When do we divide basically?


Thanks,
Megha.

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Re: M26-13   [#permalink] 01 Mar 2019, 01:38
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