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A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. If each block measures 1 by 1 meter, in how many different color patterns can the floor be parqueted? A. 104 B. 213 C. 577 D. 705 E. 726
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16 Sep 2014, 00:58
Official Solution:A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. If each block measures 1 by 1 meter, in how many different color patterns can the floor be parqueted? A. 104 B. 213 C. 577 D. 705 E. 726 Imagine the case in which we have not 5 blocks of each color but 6, then each slot from \(2*3=6\) would have 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 6 slots would be \(3*3*3*3*3*3=3^6\); Now, what is the difference between this hypothetical case and the one in the question? As we allowed 6 blocks of each color instead of 5, then we would get 3 patterns which are impossible when we have 5 blocks of each color: all white, all red and all black. Thus we should subtract these 3 cases: \(3^63=726\). Answer: E
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04 Dec 2014, 23:34
While I understand the solution above and the reason why it is correct I cant seem to put my hand on the flaw in my logic:
If we had 6 tiles each the amount of possibilities would be 3^6
as I see it this method assumes that these patterns never happened:
A A A A A  B A A A A A  C B B B B B  A B B B B B  C C C C C C  A C C C C C  B
If one of these patterns would happen when we were counting the possibilities we would end up with (3^5)*2. As I see it there is no other possible way in which the method of counting 3^6 would have resulted in a "bad" solution. And due to that reasoning I assumed we must subtract 6 from the result of 3^6. Now, obviously this is wrong but I cant see why.
If someone could explain the flaw in my logic and how could it be fixed using the same thinking that I implanted I would appreciate it.



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Re: M1605
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05 Dec 2014, 01:02
Hi Bunuel
The solution provided by you is the fastest. I tried to get the answer with a longer approach and couldn't get the right answer. Can please suggest where am I going wrong.
Area of floor = 3*2 = 6 Area of each block = 1*1 =1 No of blocks required = 6
We have 5 blocks of each white, black and red color.
Total color wise block arrangements that are possible to get from 6 blocks = 5+1, 4+1+1, 3+2+1 and 2+2+2 total ways to get blocks in (5+1) arrangement = 3C1*2C1*6!/5! = 3*2*6 = 36 total ways to get blocks in (4+1+1) arrangement = 3C1*6!/4! = 3*6*5 = 90 total ways to get blocks in (3+2+1) arrangement = 3C1*2C1*6!/2!3! = 3*2*6*5*4*3!/2*3! = 6*6*10 = 360 total ways to get blocks in (2+2+2) arrangement = 6!/2!2!2! = 6*5*4*3*2/2*2*2 = 30*3 = 90
Total = 36+90+360+90 = 576



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05 Dec 2014, 04:54
desaichinmay22 wrote: Hi Bunuel
The solution provided by you is the fastest. I tried to get the answer with a longer approach and couldn't get the right answer. Can please suggest where am I going wrong.
Area of floor = 3*2 = 6 Area of each block = 1*1 =1 No of blocks required = 6
We have 5 blocks of each white, black and red color.
Total color wise block arrangements that are possible to get from 6 blocks = 5+1, 4+1+1, 3+2+1 and 2+2+2 total ways to get blocks in (5+1) arrangement = 3C1*2C1*6!/5! = 3*2*6 = 36 total ways to get blocks in (4+1+1) arrangement = 3C1*6!/4! = 3*6*5 = 90 total ways to get blocks in (3+2+1) arrangement = 3C1*2C1*6!/2!3! = 3*2*6*5*4*3!/2*3! = 6*6*10 = 360 total ways to get blocks in (2+2+2) arrangement = 6!/2!2!2! = 6*5*4*3*2/2*2*2 = 30*3 = 90
Total = 36+90+360+90 = 576 A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. If each block measures 1 by 1 meter, in how many different color patterns can the floor be parqueted?(A) 104 (B) 213 (C) 577 (D) 705 (E) 726 There are 5 white, 5 black, and 5 red blocks available to fill 2*3=6 slots. Following 6 cases are possible for different pattern arrangements: 51: 5 blocks of the same color and 1 block of different color: \(C^1_3*C^1_2*\frac{6!}{5!}=36\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 5 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 1 blocks, and \(\frac{6!}{5!}\) is ways of different arrangements of 6 blocks, XXXXXY, out of which 5 are identical); 42: 4 blocks of the same color and 2 block of different color: \(C^1_3*C^1_2*\frac{6!}{4!*2!}=90\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 4 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 2 blocks, and \(\frac{6!}{4!*2!}\) is ways of different arrangements of 6 blocks, XXXXYY, out of which 4 X's and 2 Y's are identical); The same way for other patterns: 411: \(C^1_3*\frac{6!}{4!}=90\); 33: \(C^2_3*\frac{6!}{3!*3!}=60\); 321: \(C^1_3*C^1_2*\frac{6!}{3!*2!}=360\); 222: \(\frac{6!}{2!*2*2!}=90\); Total: 36+90+90+60+360+90=726. Answer: E. Hope it helps.
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09 Aug 2016, 10:01
I think this is a highquality question and I agree with explanation. Great question and explanation



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26 Nov 2016, 02:38
I thought the question was asking for "number of color patterns". So I assumed WWW/BBB/RRR and WW/BB/RR patterns for 6 blocks.



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11 Feb 2017, 05:02
Hi Bunuel, I was wondering why this cannot be just 15C6...i.e select six blocks from total of 15?



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11 Feb 2017, 11:42
akhisysnl wrote: Hi Bunuel, I was wondering why this cannot be just 15C6...i.e select six blocks from total of 15? 6 out of 15 gives the number of groups of 6 possible out of 15, without arrangements of these 6 blocks. Check two different solutions given above: https://gmatclub.com/forum/m16184074.html#p1414803https://gmatclub.com/forum/m16184074.html#p1451529Hope it helps.
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Re: M1605
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05 Jul 2017, 03:01
Hi brunel
in one of the above posts while listing arrangements , one arrangement you mention was
222: 6!2!∗2∗2!=906!2!∗2∗2!=90;
why we are not considering 221 = 3c1 * 2c1 * 6!2!∗2∗2!.Ideally 1st color shold be chosen out of 3 , then 2nd color and then remaining colour.After choosing arrangements need to be follow



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Re: M1605
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03 Jan 2018, 06:26
Hi Bunuel, can you please explain the point made by Blackwhite above? While I understand the solution above and the reason why it is correct I cant seem to put my hand on the flaw in my logic: If we had 6 tiles each the amount of possibilities would be 3^6 as I see it this method assumes that these patterns never happened: A A A A A  B A A A A A  C B B B B B  A B B B B B  C C C C C C  A C C C C C  B If one of these patterns would happen when we were counting the possibilities we would end up with (3^5)*2. As I see it there is no other possible way in which the method of counting 3^6 would have resulted in a "bad" solution. And due to that reasoning I assumed we must subtract 6 from the result of 3^6. Now, obviously this is wrong but I cant see why. If someone could explain the flaw in my logic and how could it be fixed using the same thinking that I implanted I would appreciate it.



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Re: M1605
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19 Apr 2018, 04:11
Easiest way to solve this problem:
The solution has to be a multiple of 3x2=6
The only option which is divisible by 6 is (E) 726/6 = 121



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07 Feb 2019, 20:09
Why doesn't 15!/(6!*9!) work as a solution?



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soco9076 wrote: Why doesn't 15!/(6!*9!) work as a solution? I had the same question! Bunuel



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21 Feb 2019, 02:18
Bunuel wrote: desaichinmay22 wrote: Hi Bunuel
The solution provided by you is the fastest. I tried to get the answer with a longer approach and couldn't get the right answer. Can please suggest where am I going wrong.
Area of floor = 3*2 = 6 Area of each block = 1*1 =1 No of blocks required = 6
We have 5 blocks of each white, black and red color.
Total color wise block arrangements that are possible to get from 6 blocks = 5+1, 4+1+1, 3+2+1 and 2+2+2 total ways to get blocks in (5+1) arrangement = 3C1*2C1*6!/5! = 3*2*6 = 36 total ways to get blocks in (4+1+1) arrangement = 3C1*6!/4! = 3*6*5 = 90 total ways to get blocks in (3+2+1) arrangement = 3C1*2C1*6!/2!3! = 3*2*6*5*4*3!/2*3! = 6*6*10 = 360 total ways to get blocks in (2+2+2) arrangement = 6!/2!2!2! = 6*5*4*3*2/2*2*2 = 30*3 = 90
Total = 36+90+360+90 = 576 A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. If each block measures 1 by 1 meter, in how many different color patterns can the floor be parqueted?(A) 104 (B) 213 (C) 577 (D) 705 (E) 726 There are 5 white, 5 black, and 5 red blocks available to fill 2*3=6 slots. Following 6 cases are possible for different pattern arrangements: 51: 5 blocks of the same color and 1 block of different color: \(C^1_3*C^1_2*\frac{6!}{5!}=36\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 5 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 1 blocks, and \(\frac{6!}{5!}\) is ways of different arrangements of 6 blocks, XXXXXY, out of which 5 are identical); 42: 4 blocks of the same color and 2 block of different color: \( C^1_3*C^1_2*\frac{6!}{4!*2!}=90\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 4 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 2 blocks, and \(\frac{6!}{4!*2!}\) is ways of different arrangements of 6 blocks, XXXXYY, out of which 4 X's and 2 Y's are identical); The same way for other patterns: 411: \(C^1_3*\frac{6!}{4!}=90\); 33: \( C^2_3*\frac{6!}{3!*3!}=60\); 321: \(C^1_3*C^1_2*\frac{6!}{3!*2!}=360\); 222: \(\frac{6!}{2!*2*2!}=90\); Total: 36+90+90+60+360+90=726. Answer: E. Hope it helps. Hi Bunuel, For your combinations 42 and 33, why did you do 3c1 and 2c1 for the 42 combination, but 3c2 for the 33 combination. I'm thinking it should be the same because in both cases you are selecting two colors right? (3B and 3R vs 4B and 2R), either way it should be 3c2?



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Re: M1605
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01 May 2019, 05:43
Bunuel  Can you please elaborate on what you wrote ? I am somehow unable to understand why cannot the solution be just 15C6. Bunuel wrote: akhisysnl wrote: Hi Bunuel, I was wondering why this cannot be just 15C6...i.e select six blocks from total of 15? 6 out of 15 gives the number of groups of 6 possible out of 15, without arrangements of these 6 blocks. Check two different solutions given above: https://gmatclub.com/forum/m16184074.html#p1414803https://gmatclub.com/forum/m16184074.html#p1451529Hope it helps.



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14 Jun 2019, 01:47
NinetyFour wrote: Bunuel wrote: desaichinmay22 wrote: Hi Bunuel
The solution provided by you is the fastest. I tried to get the answer with a longer approach and couldn't get the right answer. Can please suggest where am I going wrong.
Area of floor = 3*2 = 6 Area of each block = 1*1 =1 No of blocks required = 6
We have 5 blocks of each white, black and red color.
Total color wise block arrangements that are possible to get from 6 blocks = 5+1, 4+1+1, 3+2+1 and 2+2+2 total ways to get blocks in (5+1) arrangement = 3C1*2C1*6!/5! = 3*2*6 = 36 total ways to get blocks in (4+1+1) arrangement = 3C1*6!/4! = 3*6*5 = 90 total ways to get blocks in (3+2+1) arrangement = 3C1*2C1*6!/2!3! = 3*2*6*5*4*3!/2*3! = 6*6*10 = 360 total ways to get blocks in (2+2+2) arrangement = 6!/2!2!2! = 6*5*4*3*2/2*2*2 = 30*3 = 90
Total = 36+90+360+90 = 576 A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. If each block measures 1 by 1 meter, in how many different color patterns can the floor be parqueted?(A) 104 (B) 213 (C) 577 (D) 705 (E) 726 There are 5 white, 5 black, and 5 red blocks available to fill 2*3=6 slots. Following 6 cases are possible for different pattern arrangements: 51: 5 blocks of the same color and 1 block of different color: \(C^1_3*C^1_2*\frac{6!}{5!}=36\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 5 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 1 blocks, and \(\frac{6!}{5!}\) is ways of different arrangements of 6 blocks, XXXXXY, out of which 5 are identical); 42: 4 blocks of the same color and 2 block of different color: \( C^1_3*C^1_2*\frac{6!}{4!*2!}=90\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 4 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 2 blocks, and \(\frac{6!}{4!*2!}\) is ways of different arrangements of 6 blocks, XXXXYY, out of which 4 X's and 2 Y's are identical); The same way for other patterns: 411: \(C^1_3*\frac{6!}{4!}=90\); 33: \( C^2_3*\frac{6!}{3!*3!}=60\); 321: \(C^1_3*C^1_2*\frac{6!}{3!*2!}=360\); 222: \(\frac{6!}{2!*2*2!}=90\); Total: 36+90+90+60+360+90=726. Answer: E. Hope it helps. Hi Bunuel, For your combinations 42 and 33, why did you do 3c1 and 2c1 for the 42 combination, but 3c2 for the 33 combination. I'm thinking it should be the same because in both cases you are selecting two colors right? (3B and 3R vs 4B and 2R), either way it should be 3c2? Hi BunuelCan you please clarify point raised by NinetyFourThanks










