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M16-05

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Bunuel
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M16-05 [#permalink] New post   16 Sep 2014, 00:58
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Question Stats:

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A rectangular floor has dimensions of 2 meters by 3 meters. There are 5 white, 5 black, and 5 red parquet tiles available, each measuring 1 meter by 1 meter. How many unique color patterns can be created by arranging these tiles to cover the floor?

A. 104
B. 213
C. 577
D. 705
E. 726
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E
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Bunuel
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M16-05 [#permalink] New post   05 Dec 2014, 04:54
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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

Alternatively, we can approach the problem as follows:

There are 5 white, 5 black, and 5 red blocks available to fill the 6 slots in the 2x3 meter floor. The following 6 cases are possible for different pattern arrangements:

    5-1: 5 blocks of the same color and 1 block of a different color: \(C^1_3 * C^1_2 * \frac{6!}{5!} = 36\), (where \(C^1_3\) represents the ways to choose 1 color from 3, which will provide us with 5 blocks, \(C^1_2\) represents the ways to choose 1 color from the 2 remaining colors, which will provide us with 1 block, and \(\frac{6!}{5!}\) is the number of different arrangements of the 6 blocks, XXXXXY, out of which 5 are identical);

    4-2: 4 blocks of the same color and 2 blocks of a different color: \(C^1_3 * C^1_2 * \frac{6!}{4! * 2!} = 90\), (where \(C^1_3\) represents the ways to choose 1 color from 3, which will provide us with 4 blocks, \(C^1_2\) represents the ways to choose 1 color from the 2 remaining colors, which will provide us with 2 blocks, and \(\frac{6!}{4! * 2!}\) is the number of different arrangements of the 6 blocks, XXXXYY, out of which 4 X's and 2 Y's are identical);

Similarly, 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: \(36 + 90 + 90 + 60 + 360 + 90 = 726\).

Answer: E.

Hope it helps.
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Re M16-05 [#permalink] New post   16 Sep 2014, 00:58
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Official Solution:

A rectangular floor has dimensions of 2 meters by 3 meters. There are 5 white, 5 black, and 5 red parquet tiles available, each measuring 1 meter by 1 meter. How many unique color patterns can be created by arranging these tiles to cover the floor?

A. 104
B. 213
C. 577
D. 705
E. 726


First, let's note that since the dimensions of the floor are 2 meters by 3 meters, we need to use 6 tiles, each measuring 1 meter by 1 meter, to cover the entire floor.

Let's consider a hypothetical scenario where we have 6 blocks of each color instead of 5. In this case, each of the 6 slots in the 2x3 meter floor can be filled with one of the three colors: white, black, or red. This results in a total of \(3^6\) different ways to arrange the tiles on the floor.

Now, let's examine the differences between this hypothetical case and the actual problem where we have only 5 blocks of each color. In the hypothetical case, we can create three patterns that are impossible with only 5 blocks of each color: all white, all red, and all black. To account for this, we need to subtract these 3 cases from the total arrangements calculated earlier: \(3^6 - 3 = 726\).

Therefore, there are 726 unique color patterns that can be created using the available parquet tiles to cover the 2x3 meter floor.


Answer: E
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desaichinmay22
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Re: M16-05 [#permalink] New post   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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Re: M16-05 [#permalink] New post   09 Aug 2016, 10:01
I think this is a high-quality question and I agree with explanation. Great question and explanation
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Re: M16-05 [#permalink] New post   11 Feb 2017, 05:02
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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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Re: M16-05 [#permalink] New post   11 Feb 2017, 11:42
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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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Re: M16-05 [#permalink] New post   19 Apr 2018, 04:11
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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
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Re: M16-05 [#permalink] New post   28 Jul 2019, 11:30
Hello VeritasKarishma Bunuel
Why is calculating (15!*6!)/(9!*6!) not giving me the same answer?
(I've multiplied 6! in the numerator for the arrangement.)

Please answer! Thank you in advance!!
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Re: M16-05 [#permalink] New post   30 Jul 2019, 03:19
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hbawa993 wrote:
Hello VeritasKarishma Bunuel
Why is calculating (15!*6!)/(9!*6!) not giving me the same answer?
(I've multiplied 6! in the numerator for the arrangement.)

Please answer! Thank you in advance!!


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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Re: M16-05 [#permalink] New post   11 Oct 2019, 13:49
I think this is a high-quality question and I agree with explanation.
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Re: M16-05 [#permalink] New post   23 Oct 2020, 21:21
I think this is a high-quality question and I agree with explanation.
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Re: M16-05 [#permalink] New post   14 Jun 2021, 08:46
Bunuel VeritasKarishma
This is high quality question indeed, but the explanation isn't clear.
Using the basics of Combinatorics, we have to arrange 15 boxes in 6 positions.
So, it can be written as 15P6, where P stands for Permutations.
The expression can be written as 15! / (15-6)! or 15! / 9!.
Now the resultant is to be divided by 5! * 5! * 5!, because we have 5 identical items for all three colours.
.
What's wrong in this approach ?
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Re: M16-05 [#permalink] New post   14 Jun 2021, 21:31
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BankerBro wrote:
Bunuel VeritasKarishma
This is high quality question indeed, but the explanation isn't clear.
Using the basics of Combinatorics, we have to arrange 15 boxes in 6 positions.
So, it can be written as 15P6, where P stands for Permutations.
The expression can be written as 15! / (15-6)! or 15! / 9!.
Now the resultant is to be divided by 5! * 5! * 5!, because we have 5 identical items for all three colours.
.
What's wrong in this approach ?


This is not correct.
15P6 is "select 6 elements from 15 distinct elements and arrange the 6 selected"

Here, we do not have 15 distinct elements. 5 are white (identical), 5 red (identical) and 5 black (identical).
But note that each of the 6 block spaces on the floor are distinct.

For each spot, there are 3 ways (white, red or black) giving us 3*3*3*3*3*3 = 729 ways
But in 3 of these, all blocks are of the same colour (All 6 blocks are red or all are white or all are black). We need to remove these because we have only 5 of each block type.
Hence we get 726 total arrangements.
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Re: M16-05 [#permalink] New post   06 Jul 2022, 10:40
I think this is a high-quality question and I agree with explanation.
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Re: M16-05 [#permalink] New post   25 Apr 2023, 02:25
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M16-05 [#permalink] New post   21 Sep 2023, 03:02
I think this is a high-quality question and I agree with explanation. I think this is a high-quality question.

I also come up with the idea of 3^6 in the hypothetical situation but I couldn't think of how to minus the cases in real situation of 5-5-5 each
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Re: M16-05 [#permalink] New post   24 Apr 2024, 01:37
KarishmaB

Quote:
Here, we do not have 15 distinct elements. 5 are white (identical), 5 red (identical) and 5 black (identical).
But note that each of the 6 block spaces on the floor are distinct.

For each spot, there are 3 ways (white, red or black) giving us 3*3*3*3*3*3 = 729 ways
But in 3 of these, all blocks are of the same colour (All 6 blocks are red or all are white or all are black). We need to remove these because we have only 5 of each block type.
Hence we get 726 total arrangements.


Thank you for your clear explanation, the best explaination imo out of all the other solutions provided. Yours is simple and understandable in a very easy way.

The other solutions are appreciated as well but too complex!

Cheers!
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Re: M16-05 [#permalink]
24 Apr 2024, 01:37
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