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• The winning strategy for a high GRE score

January 17, 2019

January 17, 2019

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Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL.

M09-20

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Math Expert
Joined: 02 Sep 2009
Posts: 52108

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15 Sep 2014, 23:40
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Difficulty:

45% (medium)

Question Stats:

62% (00:39) correct 38% (00:55) wrong based on 104 sessions

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There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB, how many color patterns are possible?

A. 10
B. 12
C. 24
D. 60
E. 100

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Math Expert
Joined: 02 Sep 2009
Posts: 52108

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15 Sep 2014, 23:40
Official Solution:

There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB, how many color patterns are possible?

A. 10
B. 12
C. 24
D. 60
E. 100

THEORY

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:
$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$

For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

BACK TO QUESTION

Number of permutations of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

So, for our case the number of permutations of 5 letters BBRRR of which 2 B's and 3 R's are identical is $$\frac{5!}{2!*3!}=10$$.

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Current Student
Joined: 23 Nov 2016
Posts: 75
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51

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02 Mar 2017, 18:25
Bunuel wrote:
Official Solution:

There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB, how many color patterns are possible?

A. 10
B. 12
C. 24
D. 60
E. 100

THEORY

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:
$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$

For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

BACK TO QUESTION

Number of permutations of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

So, for our case the number of permutations of 5 letters BBRRR of which 2 B's and 3 R's are identical is $$\frac{5!}{2!*3!}=10$$.

Hey Bunuel, I think you provided the solution to a different question here .
Math Expert
Joined: 02 Sep 2009
Posts: 52108

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03 Mar 2017, 00:51
brooklyndude wrote:
Bunuel wrote:
Official Solution:

There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB, how many color patterns are possible?

A. 10
B. 12
C. 24
D. 60
E. 100

THEORY

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:
$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$

For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

BACK TO QUESTION

Number of permutations of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

So, for our case the number of permutations of 5 letters BBRRR of which 2 B's and 3 R's are identical is $$\frac{5!}{2!*3!}=10$$.

Hey Bunuel, I think you provided the solution to a different question here .

Why?

There are 3 red chips and 2 blue chips, we have RRRBB: the number of permutations of 5 letters BBRRR of which 2 B's and 3 R's are identical is $$\frac{5!}{2!*3!}=10$$.
_________________
Current Student
Joined: 28 Dec 2016
Posts: 88
Location: United States (IL)
Concentration: Marketing, General Management
Schools: Johnson '20 (M)
GMAT 1: 700 Q47 V38

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02 Apr 2017, 10:39
FYI,

BACK TO QUESTION

Number of permutations of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!/4!3!2!

.
Re: M09-20 &nbs [#permalink] 02 Apr 2017, 10:39
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