voodoochild wrote:
Bunuel,
Can you please explain why you chose 3C2 for "3-3" and 3C1*2C1 for "4-2" set above?
Similar for three color set, I am not sure why you chose 1C1 * 1C1 * 1C1 for 2-2-2 and 3C1 * 2C1 for 3-2-1 set. above. For 3-2-1, 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 2-2-2 , 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 out-of-control 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
3-3 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 3-3 case combinations = (3C2)*(6C3) = 3*20 =
60.
Now, the
4-2 case. This is very tricky. In the 3-3 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 4-2 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
2-2-2 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 2-2-2 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
3-2-1 case, yes, we will choose three colors again, but now,
unlike the 2-2-2 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 3-2-1 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 iron-man 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/gmat-quant-how-to-count/Let me know if anyone reading this has any further questions.
Mike
_________________
Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)