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Math Expert V
Joined: 02 Sep 2009
Posts: 59712

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23 00:00

Difficulty:   95% (hard)

Question Stats: 33% (02:22) correct 67% (02:42) wrong based on 54 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

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Math Expert V
Joined: 02 Sep 2009
Posts: 59712

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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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Manager  Joined: 21 Sep 2012
Posts: 206
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 59712

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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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Current Student Joined: 29 Jan 2013
Posts: 40
Location: United States
Schools: Booth PT '20 (M)
GMAT 1: 650 Q50 V26 WE: Manufacturing and Production (Manufacturing)

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I think this is a high-quality question and I agree with explanation. Great question and explanation
Intern  B
Joined: 26 Nov 2013
Posts: 2
Schools: ISB '18, IIMA , IIMB

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Hi Bunuel,
I was wondering why this cannot be just 15C6...i.e select six blocks from total of 15?
Math Expert V
Joined: 02 Sep 2009
Posts: 59712

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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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Intern  B
Joined: 10 Jun 2017
Posts: 23

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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
Intern  B
Joined: 28 Sep 2016
Posts: 3

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Why is calculating (15!*6!)/(9!*6!) not giving me the same answer?
(I've multiplied 6! in the numerator for the arrangement.)

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9874
Location: Pune, India

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hbawa993 wrote:
Why is calculating (15!*6!)/(9!*6!) not giving me the same answer?
(I've multiplied 6! in the numerator for the arrangement.)

15! means you are arranging 15 distinct things in 15 spots. Dividing this by 9! means that 9 of these 15 things are identical. You are multiplying and dividing by 6! which just cancels them off.
So your expression is the number of ways in which you can arrange say, ABCDEFGGGGGGGGG.
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Karishma
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I think this is a high-quality question and I agree with explanation. Re M16-05   [#permalink] 11 Oct 2019, 13:49
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# M16-05

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