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Choices we make!! Experts pls help

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Choices we make!! Experts pls help

by mj78ind » Mon Aug 16, 2010 8:27 pm
1. 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? (could not solve)
A. 104
B. 213
C. 577
D. 705
E. 726

I am not sure if the OA is right or the method but here it is:
[spoiler]You have to fill up 6 blocks, and each block could be white, black, or red, a total of 3 picks.
So, 6 blocks could have a total of 3^6, or 729
different combinations. Since we only have 5 blocks of each color, you have to rule out the combinations of all 6 being white, black or red, so 729 - 3 = 726.[/spoiler]

My question is why do we not do 15C6 ?? (out of 15 tiles we have to select 6), is it because there are only 3 types of tiles .... some more guidance on when to use the nCr formula vs n^r approach will be much appreciated.

Thanks!
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by kvcpk » Mon Aug 16, 2010 11:35 pm
Combinations formula doesnt work here directly. Because, we are looking for different patterns that can be formed.

you need to put 6 blocks and see how many different patterns can be formed.

for 1st block - 3 colors are available. Hence 3 patterns (R,B,W)
when 2nd block is placed, we can create 3^2 = 9 patterns
(RR,RB,RW,BB,BR,BW,WW,WR,WB)
similarly, when 3rd block is placed - 3^3 patterns
...
..
..
when 6th block is placed - 3^6 patterns [assuming that there are 6 blocks of each]

Now, 3 patterns[in which all 6 are white, all 6 are black, all 6 are red] are to be removed because we have only 5 blocks of each.

Hence 3^6 -3 = 726.

Hope this helps!!
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by tomada » Thu Aug 19, 2010 3:28 pm
kvcpk, that's a great way to solve the problem. I went crazy with combinations and, while I arrived at the same answer, it took me much longer to solve it.

kvcpk wrote:Combinations formula doesnt work here directly. Because, we are looking for different patterns that can be formed.

you need to put 6 blocks and see how many different patterns can be formed.

for 1st block - 3 colors are available. Hence 3 patterns (R,B,W)
when 2nd block is placed, we can create 3^2 = 9 patterns
(RR,RB,RW,BB,BR,BW,WW,WR,WB)
similarly, when 3rd block is placed - 3^3 patterns
...
..
..
when 6th block is placed - 3^6 patterns [assuming that there are 6 blocks of each]

Now, 3 patterns[in which all 6 are white, all 6 are black, all 6 are red] are to be removed because we have only 5 blocks of each.

Hence 3^6 -3 = 726.

Hope this helps!!
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by uwhusky » Thu Aug 19, 2010 3:42 pm
My shortcut to answer this question is:

Eni Meeni Mini Mo, pokemon, I choose you!
Yep.
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by Rahul@gurome » Thu Aug 19, 2010 5:13 pm
Solution:
Try to understand this problem through a smaller example.
Suppose there are 5 A's and 5 B's and there are 2 places into which they can be put so that there is a different arrangement each time.

So the possibilities are A A
A B
B A
B B

They are 4 in number.
This can be derived in another way.
The first place can have 2 things, either A or B.
The second place can also have 2 things, either A or B.
So total number of possibilities is 2 * 2 = 4.

Similarly take another problem.
Suppose there are 2 A's, 5 B's and and there are 3 places in which they can be arranged.
So the possibilities are A A B, A B A, B A A, A B B, B A B, B B A, B B B.
These are 7 in number.
The answer can be derived in another way.
First place can have 2 things,A or B.
Second place can have 2 things, A or B.
Third place can also have 2 things, A or B.
So the possible number of arrangements are 2*2*2 = 8.
But we have to omit 1 arrangement where there are 3 A's that is A A A because we have only 2 A's given.
So the final answer is 8 - 1 = 7

For this particular problem there are 6 places and 5 white, 5 black and 5 red blocks.
Or the possible number of arrangements is 3*3*3*3*3*3 - 3 = 3^6 - 3 = 726.

The 3 arrangements that are subtracted are the ones where there are 6 white blocks, 6 black blocks and 6 red blocks because they are all 5 in number.
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by limestone » Fri Aug 20, 2010 12:40 am
I just wanna extend the topic a little more. How many sets (combination) of block can be formed from the above topic? I try to solve it and got the number 25, and this is how I get it:
There are six slots on the floor, with maximum of 5 blocks in each color R,W,B
1st case: 5R, 1 slot left, may be 1 W or 0 W (it means 1 B) : 2 sets
2nd case: 4R, 2 slots left : WW, WB, or BB : 3 sets
3rd : 3R, 3 slots left : WWW, WWB, WBB, BBB 4 sets
4th : 2R, 5 sets (the rule is the number of slot left plus 1)
5th : 1R, 6 sets
6th : NO R : WWWWWB, WWWWBB, WWWBBB, WWBBBB,WBBBBB ( not apply the rule due to not have enough 6 blocks of W or B) : 5 sets

Total sets: 2+3+4+5+6+5 = 25
Is my approach correct? Will appreciate so much if someone can give a more concise approach or a general rule for this.
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by ILULA08 » Thu Feb 18, 2016 7:11 am
I was trying to solve this and I just have one question

Instead of using 3^6 -3 = 726 (which totally makes sense)

why is 3^5 * 2 ( first 5 picks have 3 options each and the last 6th pick has 2 of the remaining options) is wrong. I know its wrong but why? Thanks!!
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by Brent@GMATPrepNow » Thu Feb 18, 2016 8:04 am
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 six spots on the 2x3 floor. We'll label them as #1, 2, 3, 4, 5, and 6.
Take the task of placing a block in each spot and break it into stages.

Stage 1: Select a colored block for space #1
There are 3 colors to choose from, so we can complete stage 1 in 3 ways

Stage 2: Select a colored block for space #2
There are 3 colors to choose from, so we can complete stage 2 in 3 ways

Stage 3: Select a colored block for space #3
There are 3 colors to choose from, so we can complete stage 3 in 3 ways
.
.
.
Stage 6: Select a colored block for space #6
There are 3 colors to choose from, so we can complete stage 6 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 6 stages (and thus place 6 blocks) in (3)(3)(3)(3)(3)(3) ways (= 729 ways)

IMPORTANT: This method allows for the possibility that all 6 blocks being the same color. However, since there are only 5 blocks of each color, we can't have all 6 blocks the same color.

So, we need to subtract from 729 all of the arrangements where the 6 blocks are the same color.
Well, there are 3 such arrangements: 1) all blocks white, 2) all blocks black, and 3) all blocks red.

When we subtract the 3 impossible arrangements from 729, we get 726

Answer: E
----------------------
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by Matt@VeritasPrep » Thu Feb 18, 2016 6:00 pm
ILULA08 wrote:I was trying to solve this and I just have one question

Instead of using 3^6 -3 = 726 (which totally makes sense)

why is 3^5 * 2 ( first 5 picks have 3 options each and the last 6th pick has 2 of the remaining options) is wrong. I know its wrong but why? Thanks!!
There isn't a restriction on the final block. This is more like the answer to a question such as "How many arrangements are possible if not all the blocks are the same color?"
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by gocoder » Thu Jul 14, 2016 5:15 pm
I'm wondering why we can't do the following way
Unique colors *color without variation
5c1*5c1*5c1*(12c3)
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by Matt@VeritasPrep » Wed Jul 20, 2016 9:13 pm
gocoder wrote:I'm wondering why we can't do the following way
Unique colors *color without variation
5c1*5c1*5c1*(12c3)
We aren't obliged to use one of each color, so we wouldn't want to choose one of each. (We also don't know how many of each color we have, so we'd have to do some pretty messy casework to approach it that way: e.g. 1 white, 2 black, 3 red + 2 white, 2 black, 2 red + ... yuck!)
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