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M16-05

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:58
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Difficulty:

95% (hard)

Question Stats:

31% (01:39) correct 69% (02:15) wrong based on 58 sessions

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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
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43335

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15 Sep 2014, 23:58
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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: $$3^6-3=726$$.

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04 Dec 2014, 22:34
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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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05 Dec 2014, 00:02
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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Math Expert
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05 Dec 2014, 03:54
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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:

5-1: 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);

4-2: 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:
4-1-1: $$C^1_3*\frac{6!}{4!}=90$$;
3-3: $$C^2_3*\frac{6!}{3!*3!}=60$$;
3-2-1: $$C^1_3*C^1_2*\frac{6!}{3!*2!}=360$$;
2-2-2: $$\frac{6!}{2!*2*2!}=90$$;

Total: 36+90+90+60+360+90=726.

Hope it helps.
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09 Aug 2016, 09:01
I think this is a high-quality question and I agree with explanation. Great question and explanation

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26 Nov 2016, 01:38
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, 04:02
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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, 10:42
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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/m16-184074.html#p1414803
https://gmatclub.com/forum/m16-184074.html#p1451529

Hope it helps.
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05 Jul 2017, 02:01
Hi brunel

in one of the above posts while listing arrangements , one arrangement you mention was

2-2-2: 6!2!∗2∗2!=906!2!∗2∗2!=90;

why we are not considering 2-2-1 = 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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03 Jan 2018, 05:26
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.

Kudos [?]: 2 [0], given: 8

Re: M16-05   [#permalink] 03 Jan 2018, 05:26
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