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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
22
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

28% (01:46) correct 72% (02:00) wrong based on 134 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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Re M26-13  [#permalink]

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New post 16 Sep 2014, 01:25
3
6
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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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Re: M26-13  [#permalink]

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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.
Re: M26-13 &nbs [#permalink] 03 Aug 2018, 10:51
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