Ships carry 4 different colored flags (red, blue, white, black). They send messages by arranging the flags vertically on a pole. How many messages can be sent if
a) there are no restrictions
b) at least 3 blue flags and an odd number of white flags are use
c) an even number of all the flags are used
a) is obviously 4^12, but I cannot seem to figure out the other two
how many flags can be placed on the pole at a time?
I completely apologize 12
b) (12C3)*4^9
sorry definitely wrong
You don't have enough info in your question. How many flags total fit on the pole?
The pole fits 12 vertical flags
For C it's just 4^6. Just think of every 2 flags as 1 possible color.
and for b it's 3^4
nevermind the answer to b that's wrong
i think you can go this using polynomials/power series but it's been too long since i've had a combinatorics class.
*do
b) 2^8+2^6+2^4+2^2+2^0
I get 5109380 for b)
I can only get this by considering each possible combination of whites/blues and adding. For example, find the possible flag combinations that contain 3 blues and 1 white -- (12C3)*(9C1)*2^8 -- and add to this the combinations with 3 blues and 3 whites -- (12C3)*(9C3)*2^6 -- and continue on. You have to find 25 different terms:
3Blue and 1W, 3W, 5W, 7W, 9W
4Blue and 1W, 3W, 5W, 7W
5Blue and 1W, 3W, 5W, 7W
6Blue and 1W, 3W, 5W
7Blue and 1W, 3W, 5W
8Blue and 1W, 3W
9Blue and 1W, 3W
10Blue and 1W
11Blue and 1W
It's pretty inefficient, but I haven't yet figured out a better way to do it. I have a hunch that you can find this with less computation.
Another poster pointed out that some polynomial identy like these might be helpful for summing up the series that the last poster listed:
Of course, the easiest way might just be to write a computer program. 4^12 ~ 10^7, which might be doable for a low-level language like C. Monte Carlo could give you a good estimate at least.
I'm getting pretty annoyed by these dodge ads...
Excel can easily handle that
dgasdfdjgsd wrote:
Of course, the easiest way might just be to write a computer program. 4^12 ~ 10^7, which might be doable for a low-level language like C. Monte Carlo could give you a good estimate at least.
I'm getting pretty annoyed by these dodge ads...
dgasdfdjgsd wrote:
I'm getting pretty annoyed by these dodge ads...
Firefox with adblock+ can help you with that
dgasdfdjgsd wrote:
Of course, the easiest way might just be to write a computer program. 4^12 ~ 10^7, which might be doable for a low-level language like C. Monte Carlo could give you a good estimate at least.
I think that would only take longer. After I wrote out the terms it only took a minute or two to work out on a graphing calculator, which has a function for calculating combinations. Writing a program would only take up more time. I just think there might be a more compact way of writing it. I could be wrong, and so could my answer. Haven't had a chance to try the second problem yet.
a) Infinite- cuz you can place them at different heights on pole which mean different things.
b) Same
c) Same