Math Question #1

I agree with the previous poster. This "no restrictions" thing bothers me, there are tons of other ways to come up with signals besides the obvious order of the flags.

you guys are over-analyzing this problem. it is a simple combinatorics problem, able to be solved by the just logically "counting" the possible combination in the available slots. The problem CAN BE over-analyzed like this because of the lack of specific verbiage used by the poster.

The bottom line is the communication is by way of 4 different colors available to fill 12 slots. One would assume that there would be AT LEAST 12 flags of each color available (I say at least for those who say "what if one is lost"). The 12 slots refer to specific order (not height). Then the questions are asked pretty specifically. The problem (for those who want to be sticklers) is that when reading the problem, to fully understand "No Restrictions" and what a "restriction" might include according to the problem, they would have to read ahead and read examples of the restrictions that in section (a) there are none of.

Of course, if you do want to be a stickler and read the question literally as No Restrictions and add in there where they are relative to the height of the pole (still only 12 flags on) then you would need to know height of the pole (in inches or centimeters, whichever you choose) as well as the height of the flags and if they can be scrunched down at all.

Even with that, one could assume (from the problem) that the language derived from the flying of the 12 flags only takes into account the color and order of flags, not their position relative to each other or to the pole.

For part b) the solution is as follows:

12C3(9C1*3^8+9C3*3^6+9C5*3^4+9C7*3^2+9C9*3^0)

Explanation:

The 12C3 represents the selection of blue flags. You do not care whether or not there are more than 3, you just have to have at least 3. This is factored out of each term. In each term, the combinations represent when you are choosing 1,3,5,7,and 9 flags to be white. Then, the power that the 3 is raised to in each term is the number of remaining flags that can be any color other than white.

For part c) the solution is as follows:

4^6

Explanation:

This one is much easier than part b. Think if it this way, if you have one spot filled by a flag, there must be another spot with a flag of the same color. Because of this, you are essentially elimating half of the available spots.

I am pretty certain that these are correct. The only hesitation I have is that I may have overcounted on part b)

somerandomguy wrote:

For part b) the solution is as follows:

12C3(9C1*3^8+9C3*3^6+9C5*3^4+9C7*3^2+9C9*3^0)

Explanation:

The 12C3 represents the selection of blue flags. You do not care whether or not there are more than 3, you just have to have at least 3. This is factored out of each term. In each term, the combinations represent when you are choosing 1,3,5,7,and 9 flags to be white. Then, the power that the 3 is raised to in each term is the number of remaining flags that can be any color other than white.

I am pretty certain that these are correct. The only hesitation I have is that I may have overcounted on part b)

I think you have overcounted for part b). Your answer was the one I was originally going to type, but this turns out to be too many combinations because of the repeat of blue flags in the 3^p terms.

For example, from the 12C3 combinations, two possible selections of blue flags include:

B,B,B,_,_,_,_,_,_,_,_,_

and

_,B,B,B,_,_,_,_,_,_,_,_

For the specific case of, say, 2 more blues and 7 whites, the following arrangement, would be counted several times by your method:

B,B,B,B,B,W,W,W,W,W,W,W

If you are a student, do your own work.

here is a cool site: cramster.com

somerandomguy wrote:

For part c) the solution is as follows:

4^6

Explanation:

This one is much easier than part b. Think if it this way, if you have one spot filled by a flag, there must be another spot with a flag of the same color. Because of this, you are essentially elimating half of the available spots.

I am pretty certain that these are correct. The only hesitation I have is that I may have overcounted on part b)

This is definitely wrong. Say we had 4 spots and 2 colors, A and B. You would say it's 2^2. But that's wrong, there are 8:

ABAB

BABA

AABB

BBAA

BAAB

ABBA

AAAA

BBBB

Key question* Does order matter? If R-B-BL-W = B-BL-W-R. than the question is pretty easy, and you should just do it....If not, it's quite a bit tougher, especially B.

Also, specify with C. An even number of flags as in, each value is even. Or each value is the same, that would be 1 assuming order doesn't matter. 4 factorial if it does...

Are the ships facing east-west or north-south? because if they are facing east-west the boat will be facing into the prevailing westerlies and the front pole will block the wind from the other poles, thus not giving the flags any loft and rendering them not visible

Thank you for pointing out my mistakes. It's been awhile since taking a probability course.

The answer is 4.