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M09 #20

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M09 #20 [#permalink] New post 10 Nov 2008, 02:24
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There are 3 red chips and 2 blue chips. When arranged in a row, they form a certain color pattern, for example RBRRB. How many color patterns are possible?

(A) 10
(B) 12
(C) 24
(D) 60
(E) 100

[Reveal] Spoiler: OA
A

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I cldnt understnad the explanation.
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Re: M09 #20 [#permalink] New post 10 Nov 2008, 03:00
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is the QA A ?

There are 5 chips in total , hence 5! ways of arranging them. Out of them 3 are similar and 2 are similar, hence divide by 3! and 2! . So the final outcome will be 5! / ( 3! * 2! ) = 10
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Re: M09 #20 [#permalink] New post 10 Jun 2010, 07:24
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Permutations with 2 identical groups: R(3) and B(2).
Total number of arrangements: 5!/(3!)(2!) = 10
No controversy about OA (10).
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Re: M09 #20 [#permalink] New post 07 Aug 2010, 15:29
I too did not understand the solution as to why do we need to divide by 3! and 2!. Shouldn't the answer be 5!?
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Re: M09 #20 [#permalink] New post 09 Jun 2011, 08:42
The pattern depends where the 2 blue chips will be placed in the row (the others will inevitably be red). To place the first blue chip, we have 5 possibilities, for the second one 4, so this would give 5x4=20 possibilities. But there is no difference in pattern regarding the two blue chips (they are identical), therefore, we should divide by 2, so there are 5x4/2=10 different possible patterns. The correct answer is A.
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Re: M09 #20 [#permalink] New post 10 Jun 2011, 02:24
its A
good one.
5!/3!*2!..... because 3 items and 2 items are similar.
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Re: M09 #20 [#permalink] New post 10 Jun 2011, 06:30
Why can't we just do
2*2*2*2*2 = 10?

My reasoning is that you have two choices (R/B) for 5 slots..

Am I wrong?
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Re: M09 #20 [#permalink] New post 10 Jun 2011, 06:39
As unique patterns are to be found, avoiding repeats of similar patterns, use Combinations.

Total = 5; Red=2; Blue = 3.

Finding different pattern, for 2 red chips among the total 5 chips will be the sames as the patterns formed by 3 blue chips. Which is,

5C2 = 5C3 = 5!/(3! * 2!) = 10
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Re: M09 #20 [#permalink] New post 10 Jun 2011, 06:55
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bunnbear1980 wrote:
Why can't we just do
2*2*2*2*2 = 10?

My reasoning is that you have two choices (R/B) for 5 slots..

Am I wrong?


2*2*2*2*2=32. not 10.

Your reasoning is flawed. Not in every position, we will have 2 options. What if blues are consumed in first three.

Here, we have 3reds, 2blues:
Thus, total=5!
But, we need to divide it by repetitions.
3reds: 3!
2blues: 2!
5!/3!2!=10.

Please refer "MGMAT strategy guide on Word Problems/Counting" OR "Veritas Prep guide on Combinatorics and Probability" for details.
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Re: M09 #20 [#permalink] New post 10 Jun 2011, 09:50
thanks fluke. that was a mistake on my part.. i meant 2 choices for each slot, ie 5x2=10. but i get that the choices decrease as the r/b balls are used up.. thanks so much!!
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Re: M09 #20 [#permalink] New post 10 Jun 2011, 13:37
A for me..when we identical elements, we divide the total arrangements by the factorial of the number of identical ones.!
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Re: M09 #20 [#permalink] New post 29 Oct 2011, 20:23
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A variation of this question is how many ways can you rearrange the letters in the word ASSESS:

= 6! / 4! (where 6 is the total number of letters and 4 is the total number of S's)

Also, how many ways can you rearrange the letters in the word REASSESS:

= 6! / 4! x 2! (where 2 is the total number of E's)
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Re: M09 #20 [#permalink] New post 13 Jun 2012, 04:26
Straight forward solution. This is an arrangement of 5 items involving a set of 3 identical items and another of 2 identical items.
Solution: 5!/(3!*2!)= 10 ways.

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Re: M09 #20 [#permalink] New post 13 Jun 2012, 04:52
Where the formula is giving the answer to be 10 number of ways to arrange the chip, I am getting 11 ways when arranging them manually. Those 11 ways of arrangements are as follows.
BBRRR
BRBBR
BRBRB
BRRBR
BRRRB
RBBRR
RBRBR
RBRRB
RRBBR
RRBRB
RRRBB

Can anybody help me on where am I wrong?

Last edited by doe007 on 13 Jun 2012, 05:15, edited 1 time in total.
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Re: M09 #20 [#permalink] New post 13 Jun 2012, 04:58
Expert's post
doe007 wrote:
Where the formula is giving the answer to be 10 number of ways to arrange the chip, I am getting 11 ways when arranging them manually. Those 11 ways of arrangements are as follows.
BBRRR
BRBBR
BRBRB
BRRBR
BRRRB
RBBRR
RBRBR
RBRRB
RRBBR
RRBRB
RRRBB

Can anybody help me on where am I wrong?


We have 3 R's and 2 B's.
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Re: M09 #20 [#permalink] New post 13 Jun 2012, 05:20
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ritula wrote:
There are 3 red chips and 2 blue chips. When arranged in a row, they form a certain color pattern, for example RBRRB. How many color patterns are possible?

(A) 10
(B) 12
(C) 24
(D) 60
(E) 100

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I cldnt understnad the explanation.

Hi,

This problem can be approached in two ways:
1. Total number of arrangements/(arrangements of R's & S's) = 5!/3!2! = 10
Which is basically removing the arrangements from permutations to get the combinations, or,
2. Total number of ways in which 3 R's can be arranged at 5 places, rest of the places would be filled by B,
or 5C3 = 10

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Re: M09 #20 [#permalink] New post 13 Jun 2012, 06:28
Whenever you have a question like this with two identical sets, think of it as 5! where you have a total of 5x4x3x2x1= 120 different sets of rearrangements. But you also have 3 identical red chips and 2 identical blue chips so you MUST account for these as well.

5!/3!2!= 5x4x3x2x1/ 3x2x1x2x1 which equals out to 10 different arrangements being possible.

Would this be the most efficient approach on the GMAT? :roll:
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Re: M09 #20 [#permalink] New post 13 Jun 2012, 06:35
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tisrar wrote:
Whenever you have a question like this with two identical sets, think of it as 5! where you have a total of 5x4x3x2x1= 120 different sets of rearrangements. But you also have 3 identical red chips and 2 identical blue chips so you MUST account for these as well.

5!/3!2!= 5x4x3x2x1/ 3x2x1x2x1 which equals out to 10 different arrangements being possible.

Would this be the most efficient approach on the GMAT? :roll:



THEORY FOR SUCH KIND OF PERMUTATION QUESTIONS:

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.

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

BACK TO THE ORIGINAL QUESTION:
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

According to the above the # of permutations of 5 letters BBRRR of which 2 B's and 3 R's are identical is \frac{5!}{2!*3!}=10.

Answer: A.

Hope it's clear.
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Re: M09 #20 [#permalink] New post 14 Jun 2013, 00:39
somebody asked above y would we divide by 3! and 2!.

Its because the 3 reds and 2 blues are identical on their own.
if the anagram would have been r(1)r(2)r(3)b(1)b(2) (the numbers in paranthesis is to denote the subscripts for each ) the anagram r(1)r(2)r(3)b(1)b(2) would be diff compared to r(2)r(1)r(3)b(1)b(2) .But here the case is identical .So we have to remove the redundant one's.

There fore arranging them in 5! and later dividing by 3! and 2! to remove the redundant cases.

hope it helps.
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Re: M09 #20 [#permalink] New post 23 Jun 2013, 17:12
Answer: A

The way to approach this problem is using the basic permutation/arrangement formula for n times and removing identical items:

No. of ways to arrange 5 items = 5!
No. of ways to arrange R identical items = 3!
No. of ways to arrange B identical items =2!

Hence total number of ways to arrange = 5! / (3! * 2! ) = 10
Re: M09 #20   [#permalink] 23 Jun 2013, 17:12
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