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M16 #5, Pls answer [#permalink]
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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 Source: GMAT Club Tests  hardest GMAT questions Let's assume that there are plenty of blocks of each color available and try to answer the question under this assumption. It doesn't matter whether the floor measures 2 by 3 or 6 by 1. We can always enumerate the square meters from 1 to 6. In fact, we have 6 available slots each of which can be filled with any of the three colors. There are 3 ^6 different ways of parqueting the floor. (QUESTION: Why do we use 3^6 here? Can someone pls explain this to me? Thank you.) Because there are in fact only 5 blocks of each color available, it is impossible to cover the floor with one color (this would require 6 blocks of one color). Thus, we have to exclude the three patterns that involve only one color. The final answer is 7293 = 726. .



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14 Oct 2008, 06:37
Hi all. The explanation states pretty much clearly why we use \(3^6\). I've marked the text saying that in red. Think of it in this way: You have a total of 6 blocks that have to be parquetted. You need to choose ONE of the THREE colored blocks for each of the 6 blocks. Let's imagine we have 6 blocks of each color (not 5 as it's stated in the question). Then we choose the colors for each of the blocks like this: 1st block. White, red or black (3 colors available), a total of \(3^1\) patterns 2nd block. White, red or black (3 colors available), a total of \(3^2\) patterns 3rd block. White, red or black (3 colors available), a total of \(3^3\) patterns 4th block. White, red or black (3 colors available), a total of \(3^4\) patterns 5th block. White, red or black (3 colors available), a total of \(3^5\) patterns 6th block. White, red or black (3 colors available), a total of \(3^6\) patterns But in our case we have only 5 blocks of each color available. That is why we subtract 3 patterns (all 6 white, all 6 black, all 6 red) that are not possible under these circumstances from \(3^6=729\). The answer is 726. Hope this helps. dczuchta wrote: A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. Each block measures 1 by 1 meter. In how many different color patterns can the floor be parqueted?
Let's assume that there are plenty of blocks of each color available and try to answer the question under this assumption.
It doesn't matter whether the floor measures 2 by 3 or 6 by 1. We can always enumerate the square meters from 1 to 6. In fact, we have 6 available slots each of which can be filled with any of the three colors. There are 3 ^6 different ways of parqueting the floor. (QUESTION: Why do we use 3^6 here? Can someone pls explain this to me? Thank you.)
Because there are in fact only 5 blocks of each color available, it is impossible to cover the floor with one color (this would require 6 blocks of one color). Thus, we have to exclude the three patterns that involve only one color. The final answer is 7293 = 726. .
The correct answer is E.
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25 Sep 2009, 15:28
It's not that simple. You can't use just the combinations formula here as the there are three different colors of the blocks. The number you calculated represents the number of ways you can pick 6 blocks from 15 blocks. It doesn't count all the possible patterns in it. You would be able to use that formula if you were asked for example "How many different groups of 6 can be formed from 15 people?" However, when you have a question with different objects, like blocks with different color in this question, you can't just sum 3 different sets into a single pool to pick from. Does it make sense? Can anybody explain better? kt00381n wrote: Why 15!/6!9! wont work? Im confused.
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12 Nov 2010, 12:37
HI guys, This concept is different from combinations formula. Please refer here: mathcombinatorics87345.html#p785179 read walker's first post and shrouded1's post.
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30 Dec 2010, 00:01
So each 2x3 block can be 3 possible color. 3x3x3x3x3x3=729 and subtract 3 because in the 3 situation when all 5 of a certain color is used up leaving 1 panel a different color. Nice problem.



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30 Dec 2010, 23:46
wondering if there are only 4 balls of each colour instead of 5, will the answer be 3^63^2 ??



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31 Dec 2010, 02:13
sunnydepp wrote: wondering if there are only 4 balls of each colour instead of 5, will the answer be 3^63^2 ?? In this case i think it will be 3^6  6



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13 Jan 2012, 00:29
This does not make sense. Are we not counting repeating patterns here? The question syas different pattern...
for example: The pattern BBBRRR is being counted at least a 2 times here. I s my understanding right?
Thanks!



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Jdam wrote: Tough question. I was also trapped by combinatronics method. It can be done with combinations, though this approach would be lengthier. 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. Shorter approach: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. kt00381n wrote: Why 15!/6!9! wont work? Im confused. I you look at the first approach you'll see that \(C^6_{15}\) doesn't give all possible patterns possible. \(C^6_{15}\) is # of different groups of 6 possible out of 15 distinct objects, which clearly is not the case here. diddygmat wrote: This does not make sense. Are we not counting repeating patterns here? The question syas different pattern...
for example: The pattern BBBRRR is being counted at least a 2 times here. I s my understanding right?
Thanks! Both approaches above count different patterns. Consider the following case: there are 2 slots to fill and 4 white, 4 black, and 4 red blocks available. How many different arrangements are possible: WW WB BW WR RW BB BR RB RR Total of 9 different arrangements. The same as if we consider approach #2 from above: each slot from 2 has 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 2 slots would be 3*3=3^2=9. Hope it helps.
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30 Jan 2012, 14:50
Bunuel wrote: Jdam wrote: Tough question. I was also trapped by combinatronics method. It can be done with combinations, though this approach would be lengthier. 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. Shorter approach: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. kt00381n wrote: Why 15!/6!9! wont work? Im confused. I you look at the first approach you'll see that \(C^6_15\) doesn't give all possible patterns possible. \(C^6_15\) is # of different groups of 6 possible out of 15 distinct objects, which clearly is not the case here. diddygmat wrote: This does not make sense. Are we not counting repeating patterns here? The question syas different pattern...
for example: The pattern BBBRRR is being counted at least a 2 times here. I s my understanding right?
Thanks! Both approaches above count different patterns. Consider the following case: there are 2 slots to fill and 4 white, 4 black, and 4 red blocks available. How many different arrangements are possible: WW WB BW WR RW BB BR RB RR Total of 9 different arrangements. The same as if we consider approach #2 from above: each slot from 2 has 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 2 slots would be 3*3=3^2=9. Hope it helps. Hi Bunnel, I had a question regarding the selection of colors for each case. In the first case, why are we selecting the 2 colors individually? 51: 5 blocks of the same color and 1 block of different color, why can't we have 3C2*6!/5! Regards, Anu



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30 Jan 2012, 15:37
anuu wrote: Hi Bunnel,
I had a question regarding the selection of colors for each case.
In the first case, why are we selecting the 2 colors individually?
51: 5 blocks of the same color and 1 block of different color, why can't we have 3C2*6!/5!
Regards, Anu 3C2 will give us the # of two different colors possible from 3 colors available: {WB}, {WR}, {BR}. But in our case white provides with 5 blocks and red provides with 1 block is different from white provides with 1 blocks and red provides with 5 block: {WB} and {BW} are different, so it should be the way I wrote 3C1*2C1. Hope it's clear.
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21 Jul 2012, 11:01
Bunuel wrote: 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: \(C^1_3*C^1_2*\frac{<}{span>6!/4!*2!}=90\), (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);
33: \(C^2_3*\frac{<}{span>6!/3!*3!}=60\);
Bunuel, Can you please explain why you chose 3C2 for "33" and 3C1*2C1 for "42" set above? Similar for three color set, I am not sure why you chose 1C1 * 1C1 * 1C1 for 222 and 3C1 * 2C1 for 321 set. above. For 321, we pretty much have the same restriction. The first color COULD be chosen in 3 out of 1 ways = 3C1; The second color COULD be chosen in 2C1 ways but the third color HAS to be chosen in 1C1 ways. However, for 222 , we have the same restriction, 1 out of 3 color for the first one, 1 out of 2 for the second one and 1 out of the third color. I think that I am missing something very fundamental here. Can you please explain the difference? Can you please help?



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27 Jul 2012, 10:52
voodoochild wrote: Bunuel, Can you please explain why you chose 3C2 for "33" and 3C1*2C1 for "42" set above?
Similar for three color set, I am not sure why you chose 1C1 * 1C1 * 1C1 for 222 and 3C1 * 2C1 for 321 set. above. For 321, we pretty much have the same restriction. The first color COULD be chosen in 3 out of 1 ways = 3C1; The second color COULD be chosen in 2C1 ways but the third color HAS to be chosen in 1C1 ways. However, for 222 , we have the same restriction, 1 out of 3 color for the first one, 1 out of 2 for the second one and 1 out of the third color.
I think that I am missing something very fundamental here. Can you please explain the difference? Can you please help? I'm responding to a pm from voodoochild. First of all, with all due respect, you are an outofcontrol nCr abuser  Do you get 8C1 hours of sleep at night? Do you work 8C1 hours a day, 5C1 days a week? Do you study for the GMAT 24C1/7C1/365C1? It's wildly unnecessary to write 3C1, when we can just say 3. Rather than try to interpret what Mr. Bunuel was thinking in each step, I just going to explain how I would think about this question. Let's look at the 33 case  it's a two color cases, and so there are 3C2 = 3 ways we can choose two colors. Given two colors  say red and black  then if we choose the spaces where we put the red, the spaces for the black tiles thereby will be determined. Given six spaces, in how many ways can we choose three for the red tiles? In 6C3 = 20 ways. Total number of 33 case combinations = (3C2)*(6C3) = 3*20 = 60. Now, the 42 case. This is very tricky. In the 33 case, we just had to pick two colors  say, red and black  and we were done . 3 reds and 3 blacks. Things are trickier in the 42 case, because if I pick my two colors  again, say, red and black  then I still have another choice to make: will it be 4 red and 2 blacks, or 4 blacks and 2 reds? There are 3 options on the first choice (choice of which two colors are involved) and 2 options on the second choice (which color is 4 and which one is 2). One could write that as (3C1)*(2C1), but frankly, I think that's an asinine overuse of the nCr notation. I would say: just use the Fundamental Counting Principle directly. We have three colors  first we pick one color for the 4 tiles: that's three choices; then of the remaining two colors, we pick another for the 2 tiles: that's two choices  total number of choices = 3*2 = 6. Then, how many ways can we place the 2 tiles in six spaces?  6C2 = 15. Total number of combinations = 6*15 = 90I think that same tricky thing is the root of your question in the other cases. In the 222 case, there's only one way we can have two of each color. We have to choose three colors, and there are only three colors from which to choose, so only 1 choice is possible. I think the monstrosity (1C1)*(1C1)*(1C1) should be taken out back and shot. Plain and simple, there's 1 way to choose all three colors. Done. It simply doesn't make any sense, for the 222 case, to start getting into  3 choices for the first two tiles, and 2 choices for the second pair  all that will be irrelevant, because of the symmetry. If we have 2 red, 2 white, and 2 black, that's identical to 2 white, 2 black, and 2 red. Remember, this part of the counting is about the colors chosen  we would handle the distribution of the tiles on the floor in separate step. The three possible colors are equally represented, so there's only 1 way to do that. In the 321 case, yes, we will choose three colors again, but now, unlike the 222 case, order matters. Having 3 reds, 2 blacks, and 1 white is different from having, say 3 reds, 2 whites, and 1 black. Again, I would argue that use of nCr is unnecessary, and a product of nCr's is ridiculous. Use the FCP. We have tiles of three colors, and we are going to choose for the 321 patterns. In how many ways can we choose the color that will have three tiles? Three. Given that choice, in how many ways can be pick a second color for the two tiles? Two. Once those two are determined, we have nothing else to count  the last color must have 1 tile. So, the product of the color choices is 3*2 = 6. Does all that make sense? Also, I will say about this problem  all this stuff about counting the combinations for each individual case  I guess I see why that is belabored on this page, to give readers an ironman workout in combinatorics, but breaking everything into cases is by far the long and awkward way to answer the overall question. The infinitely more elegant solution, given a few times already on this page, is as follows. We have six spaces, and each one could be one of three tiles, so that's a total of 3^6 = 729 combinations. We can make every single one of those 729 patterns except three  6 reds, or 6 blacks, or 6 whites, are the three combinations not possible, because we have only 5 of each tile. Therefore, of the 729 possibles, we can do all except three of them. 729  3 = 726. That is elegant!! In counting and combinatorics problems, there are always long and clunky ways to find the answer. The challenge, sometimes requiring considerable insight, is to find the most elegant solution. Incidentally, often that means ditching the nCr stuff and using the FCP directly. You may find this post a helpful refresher on the FCP: http://magoosh.com/gmat/2012/gmatquanthowtocount/Let me know if anyone reading this has any further questions. Mike
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27 Jul 2012, 13:45
mikemcgarry wrote: voodoochild wrote: Bunuel, Can you please explain why you chose 3C2 for "33" and 3C1*2C1 for "42" set above?
Similar for three color set, I am not sure why you chose 1C1 * 1C1 * 1C1 for 222 and 3C1 * 2C1 for 321 set. above. For 321, we pretty much have the same restriction. The first color COULD be chosen in 3 out of 1 ways = 3C1; The second color COULD be chosen in 2C1 ways but the third color HAS to be chosen in 1C1 ways. However, for 222 , we have the same restriction, 1 out of 3 color for the first one, 1 out of 2 for the second one and 1 out of the third color.
I think that I am missing something very fundamental here. Can you please explain the difference? Can you please help? I'm responding to a pm from voodoochild. First of all, with all due respect, you are an outofcontrol nCr abuser  Do you get 8C1 hours of sleep at night? Do you work 8C1 hours a day, 5C1 days a week? Do you study for the GMAT 24C1/7C1/365C1? It's wildly unnecessary to write 3C1, when we can just say 3. Rather than try to interpret what Mr. Bunuel was thinking in each step, I just going to explain how I would think about this question. Let's look at the 33 case  it's a two color cases, and so there are 3C2 = 3 ways we can choose two colors. Given two colors  say red and black  then if we choose the spaces where we put the red, the spaces for the black tiles thereby will be determined. Given six spaces, in how many ways can we choose three for the red tiles? In 6C3 = 20 ways. Total number of 33 case combinations = (3C2)*(6C3) = 3*20 = 60. Now, the 42 case. This is very tricky. In the 33 case, we just had to pick two colors  say, red and black  and we were done . 3 reds and 3 blacks. Things are trickier in the 42 case, because if I pick my two colors  again, say, red and black  then I still have another choice to make: will it be 4 red and 2 blacks, or 4 blacks and 2 reds? There are 3 options on the first choice (choice of which two colors are involved) and 2 options on the second choice (which color is 4 and which one is 2). One could write that as (3C1)*(2C1), but frankly, I think that's an asinine overuse of the nCr notation. I would say: just use the Fundamental Counting Principle directly. We have three colors  first we pick one color for the 4 tiles: that's three choices; then of the remaining two colors, we pick another for the 2 tiles: that's two choices  total number of choices = 3*2 = 6. Then, how many ways can we place the 2 tiles in six spaces?  6C2 = 15. Total number of combinations = 6*15 = 90I think that same tricky thing is the root of your question in the other cases. In the 222 case, there's only one way we can have two of each color. We have to choose three colors, and there are only three colors from which to choose, so only 1 choice is possible. I think the monstrosity (1C1)*(1C1)*(1C1) should be taken out back and shot. Plain and simple, there's 1 way to choose all three colors. Done. It simply doesn't make any sense, for the 222 case, to start getting into  3 choices for the first two tiles, and 2 choices for the second pair  all that will be irrelevant, because of the symmetry. If we have 2 red, 2 white, and 2 black, that's identical to 2 white, 2 black, and 2 red. Remember, this part of the counting is about the colors chosen  we would handle the distribution of the tiles on the floor in separate step. The three possible colors are equally represented, so there's only 1 way to do that. In the 321 case, yes, we will choose three colors again, but now, unlike the 222 case, order matters. Having 3 reds, 2 blacks, and 1 white is different from having, say 3 reds, 2 whites, and 1 black. Again, I would argue that use of nCr is unnecessary, and a product of nCr's is ridiculous. Use the FCP. We have tiles of three colors, and we are going to choose for the 321 patterns. In how many ways can we choose the color that will have three tiles? Three. Given that choice, in how many ways can be pick a second color for the two tiles? Two. Once those two are determined, we have nothing else to count  the last color must have 1 tile. So, the product of the color choices is 3*2 = 6. Does all that make sense? Also, I will say about this problem  all this stuff about counting the combinations for each individual case  I guess I see why that is belabored on this page, to give readers an ironman workout in combinatorics, but breaking everything into cases is by far the long and awkward way to answer the overall question. The infinitely more elegant solution, given a few times already on this page, is as follows. We have six spaces, and each one could be one of three tiles, so that's a total of 3^6 = 729 combinations. We can make every single one of those 729 patterns except three  6 reds, or 6 blacks, or 6 whites, are the three combinations not possible, because we have only 5 of each tile. Therefore, of the 729 possibles, we can do all except three of them. 729  3 = 726. That is elegant!! In counting and combinatorics problems, there are always long and clunky ways to find the answer. The challenge, sometimes requiring considerable insight, is to find the most elegant solution. Incidentally, often that means ditching the nCr stuff and using the FCP directly. You may find this post a helpful refresher on the FCP: http://magoosh.com/gmat/2012/gmatquanthowtocount/Let me know if anyone reading this has any further questions. Mike thanks C 1. It'sC1 clearC1 nowC1 I am sorry Mike for overusing C's.... I really liked your post.. I cannot stop laughing...haha I didn't realize that I seriously wrote this "(1C1)*(1C1)*(1C1) " But, it's funny how mind behaves sometimes....hhaha



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Re: M16 #5, Pls answer [#permalink]
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29 Jul 2012, 08:18
mikemcgarry wrote: In the 321 case, yes, we will choose three colors again, but now, unlike the 222 case, order matters. Having 3 reds, 2 blacks, and 1 white is different from having, say 3 reds, 2 whites, and 1 black. Again, I would argue that use of nCr is unnecessary, and a product of nCr's is ridiculous. Use the FCP. We have tiles of three colors, and we are going to choose for the 321 patterns. In how many ways can we choose the color that will have three tiles? Three. Given that choice, in how many ways can be pick a second color for the two tiles? Two. Once those two are determined, we have nothing else to count  the last color must have 1 tile. So, the product of the color choices is 3*2 = 6.
Mike, 2 questions: (1) I was redoing the above problem. However, for 321 case, Bunuel has calculated the number of ways to choose a color = 3C1 * 3C1 instead of 3*2. Can you please comment on that? (2) I understand that it is easier to use FCP. However, I really like Bunuel's method because it deals with the basics of combinatorics. While solving such Combinations problem, I see that you always had an eye for the order especially while solving 51 vs. 33 case. I don't think that such thoughts are mechanical. They are based on experience and real world imagination. I believe that solving above problem requires a detailed understanding of Cs and Ps + a real world imagination. The second part is the most crucial element because the only difference between 51 and 33 case is the order. ORder matters in one and not in the other. I believe that such problems cannot be solved mechanically like a robot. Correct? Thanks



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29 Jul 2012, 08:25
voodoochild wrote: mikemcgarry wrote: In the 321 case, yes, we will choose three colors again, but now, unlike the 222 case, order matters. Having 3 reds, 2 blacks, and 1 white is different from having, say 3 reds, 2 whites, and 1 black. Again, I would argue that use of nCr is unnecessary, and a product of nCr's is ridiculous. Use the FCP. We have tiles of three colors, and we are going to choose for the 321 patterns. In how many ways can we choose the color that will have three tiles? Three. Given that choice, in how many ways can be pick a second color for the two tiles? Two. Once those two are determined, we have nothing else to count  the last color must have 1 tile. So, the product of the color choices is 3*2 = 6.
Mike, 2 questions: (1) I was redoing the above problem. However, for 321 case, Bunuel has calculated the number of ways to choose a color = 3C1 * 3C1 instead of 3*2. Can you please comment on that? (2) I understand that it is easier to use FCP. However, I really like Bunuel's method because it deals with the basics of combinatorics. While solving such Combinations problem, I see that you always had an eye for the order especially while solving 51 vs. 33 case. I don't think that such thoughts are mechanical. They are based on experience and real world imagination. I believe that solving above problem requires a detailed understanding of Cs and Ps + a real world imagination. The second part is the most crucial element because the only difference between 51 and 33 case is the order. ORder matters in one and not in the other. I believe that such problems cannot be solved mechanically like a robot. Correct? Thanks In my solution it's " 321: \(C^1_3*C^1_2*\frac{6!}{3!*2!}=360\)".
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Re: M16 #5, Pls answer [#permalink]
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29 Jul 2012, 08:44
Thanks Bunuel. I think that I need to get my eyes checked.
Bunuel/Mike,
I have a side question. I see why you have written 3C1 for 411 case. It makes sense from the logical standpoint. Is there any mechanical method, apart from the logical method, that could help me in arriving at 3C1? Logically, there could be only three combinations because the first color will occupy 4 tiles. The second and the third color will occupy only one position each.
However, I am a bit confused because of three different combinations for the following three sets :
222 321 411
Each of them follows a different principle for solving the problem. The first one, as per Mike, can be solved by considering Symmetry. The second one can be solved by using FCP. I am curious  did you actually visualize these problems while solving or is there any mechanical method used to solve these problems. The reason why I am asking this question is :
1) To know whether there is any mechanical method 2) I would crash if there were, say, 8 colors and 100 tiles. I wouldn't be able to imagine so many possibilities. I know that FCP is an easier way to solve this problem. But I wanted to know whether there is any mechanical procedure that could guide us to the answer, even though I used a long method.
Appreciate your thoughts.



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07 Aug 2012, 10:14
voodoochild wrote: Mike, 2 questions: (1) I was redoing the above problem. However, for 321 case, Bunuel has calculated the number of ways to choose a color = 3C1 * 3C1 instead of 3*2. Can you please comment on that? I'm not sure where you're looking, but in Bunuel's first post on this page, for the 321 case, he has (3C1)*(2C1), same as I have. He is using a funky notation that is not what I've seen  his 3C1 has a 1 on the top and a 3 on the bottom, which is a little peculiar  but despite notational differences, the underlying math is the same as what I've done. voodoochild wrote: (2) I understand that it is easier to use FCP. However, I really like Bunuel's method because it deals with the basics of combinatorics. While solving such Combinations problem, I see that you always had an eye for the order especially while solving 51 vs. 33 case. I don't think that such thoughts are mechanical. They are based on experience and real world imagination. I believe that solving above problem requires a detailed understanding of Cs and Ps + a real world imagination. The second part is the most crucial element because the only difference between 51 and 33 case is the order. Order matters in one and not in the other. I believe that such problems cannot be solved mechanically like a robot. Correct? Correct. In fact, a blanket statement  At no time on the GMAT, whether on Quant or Verbal or AWA or IR, can you ever expect sustained success with a formulaic mechanical approach. From the time you sit down to start your GMAT until the time you stand up when you are done, your critical thinking skills must be engaged to the utmost. There is absolutely no shortcut for rigorous critical thinking skills. If you are doing anything other than exercising every facet of your intelligence to the utmost, you are not doing what the GMAT continuously demands. Does that make sense? Mike
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07 Aug 2012, 10:22
voodoochild wrote: I am curious  did you actually visualize these problems while solving or is there any mechanical method used to solve these problems. The reason why I am asking this question is: 1) To know whether there is any mechanical method As per my previous post, abandon mechanical methods, or use them at most only tentatively. Your analysis always has to include critical thinking. Here, visualizing the situation is an irreducible part of solving the problem. It very dangerous, and ultimately not helpful, to try to find a mechanical shortcut that will excuse you from doing the hard work of visualizing. voodoochild wrote: 2) I would crash if there were, say, 8 colors and 100 tiles. I wouldn't be able to imagine so many possibilities. I know that FCP is an easier way to solve this problem. But I wanted to know whether there is any mechanical procedure that could guide us to the answer, even though I used a long method. That's getting into the kind of problem that only idiotsavants who don't bathe could solve with ease. The GMAT is simply not going to ask you something of this type that is beyond what most people could visualize. Yes, there are more advanced methods for keeping track of everything when it's far more than what anyone could visualize  if you took an graduatelevel course in Combinatorics, you could learn such things. But, again, that is leagues beyond what the GMAT expects. Mike
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Re: M16 #5, Pls answer [#permalink]
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08 Aug 2012, 09:13
mikemcgarry wrote: Yes, there are more advanced methods for keeping track of everything when it's far more than what anyone could visualize  if you took an graduatelevel course in Combinatorics, you could learn such things. But, again, that is leagues beyond what the GMAT expects.
Can you please hint on this? at least the topic of the graduate level course? I am curious.....thanks




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