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There are 8 red chips and 2 blue ones. When arranged in a ro [#permalink]

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10 Oct 2009, 21:03

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There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?
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Thanks, Just think differently, there is a easier solution:)

Last edited by Bunuel on 14 Nov 2013, 12:55, edited 1 time in total.

There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?

We can ignore the red chips.

To place 2 blue chips in 10 slots we get -> 10P2 = 10!/(10-2)! = 10!/8! = 10x9 = 90

Further we need to divide this by 2 as the formula considers both the blue chips to be unique, which they are not.

There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?

We can ignore the red chips.

To place 2 blue chips in 10 slots we get -> 10P2 = 10!/(10-2)! = 10!/8! = 10x9 = 90

Further we need to divide this by 2 as the formula considers both the blue chips to be unique, which they are not.

So 45.

Hi jax91, I did this earlier but I think it is wrong. Consider this point, You are dividing the permutations by half because of arrangements of these kinds:- BBRRRRRRRR AND RRRRRRRRBB . These color patterns are the same. But there will be some arrangements which will be palindromes, such as:- BRRRRRRRRB and these will not come twice because which Blue pack is kept at the first and the last position doesn't matter. This will be true for the Blue packs coming at the following positions -> (1,10), (2,8), (3,7), (4,6) and (5,5). So, a total of 5 palindromes. Here is how I then solved the question, Case 1 - Palindromes :- As said earlier, a total of 5.

Case 2 - Number of permutations in which the chips are not equidistant from each other :- In this case you have to first consider the 10 available places made of 5 left and 5 right positions. Now, the side selected, whether left or right doesn't matter for the following example as the reason -> BB placed at (1,2) positions is same as when placed at (9,10) positions. This case has two sub-cases as follows :- 1. When both of the Blues are placed in the first or last five positions :- 5C2 = 10. (Thus, this case considers the following arrangements as the same -> BBRRRRRRRR and RRRRRRRRBB). 2. When one is placed in the left/right half and the other in the right/left half (Note: at differently distant positions) :- (5C1 X 4C1)/2 = 10. Divided by 2 as arrangements such as these are the same -> RBRRRRRRRB and BRRRRRRRBR . This gives us a total of 5+10+10 = 25 color patterns. I hope its right. Please point out any mistakes. And dont forget to Give me Some Kudos!
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There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?

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.

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 questions:

There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?

According to the above 8 red chips and 2 blue chips can be arranged in 10!/(8!2!)=45 ways.

Re: There are 8 red chips and 2 blue ones. When arranged in a ro [#permalink]

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30 Dec 2013, 07:55

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This post received KUDOS

RJAGmat wrote:

There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?

Good question. Instead of thinking of the red and blue chips as being the choices, think of the ten slots as being 10 different items (example 1 to 10) We need to choose two slots for the blue chips (remaining go to red). and the order the slots are picked doesn't matter (for example, picking slot 1 and 2 is the same as picking 2 and 1, for the blue chips)

this leaves us with a problem that involves choosing two slots out of ten choices, and since order doesn't matter, we have 10C2 (instead of 10P2) {note: I think this becomes a much easier problem to understand after you recognize that its a combinations problem with slots instead of permutations with chips, though either way isn't too difficult} 10! / (8!2!) = 45

one other way of doing this: again take the slots 1 to 10. the blues have to be in two slots. fix the first blue at one and go down the list (1,2) , (1,3) .... (1,10) = 9 instances then start at 2 (2,3) , (2,4).....(2,10) = 8 instances

this continues all the way to one. you have a total of 9 + 8 ...+1 = 45 instances, or (n) (n+1) / 2 = 9*10 / 2 = 45 {for sum 1 to n}

Re: There are 8 red chips and 2 blue ones. When arranged in a ro [#permalink]

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14 Nov 2013, 13:20

right, forgot about the exact permutuations formula, so i started working with slots instead of chips.

now that i see the formula you posted, using chips and permutations appears way simpler than switching the problem to slots and combinations, though in this case both lead you to the exact same calculation and answer (because there's only two different types in the entire range of values - you either have \(\frac{n!}{P_1! P_2!}\)for permutations of chips, or\(\frac{n!}{(n-p)!(p)!}\) for combinations of slot choices, which both lead you to \(\frac{10!}{8!2!}\) = 45)

the alternate way i posted could be used if you didn't know the permutuations or combinations formula and had to just use brute force by trying scenarios

Re: There are 8 red chips and 2 blue ones. When arranged in a ro [#permalink]

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14 Nov 2013, 14:04

Bunuel wrote:

RJAGmat wrote:

There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?

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.

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 questions:

There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?

According to the above 8 red chips and 2 blue chips can be arranged in 10!/(8!2!)=45 ways.

Hope this helps.

Bunuel, I knew this earlier but can you pls explain how to proceed if the same are not arranged in a row but in a circular way? Thank you

Re: There are 8 red chips and 2 blue ones. When arranged in a ro [#permalink]

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09 Jul 2015, 13:23

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Take 1 case: BBRRRRRRRR The right B can move 9 positions. When Left B shifts by 1 position to the right, the right B can move 8 positions, and so as left B keeps shifting, the right B can move 7,6,5,4,3,2,1 positions. So the total number of arrangements is 9+8+7+6+5+4+3+2+1=45
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There are 8 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?

Hi,

the Q can be rephrased "in 10 places for 8R and 2B, choose two places for Blue".. we can choose 2 places for blue in 10C2, combinations as both blue are same and all red are same.. \(10C2= \frac{10!}{{8!2!}}= 10*\frac{9}{2}=45\)
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