Originally posted by LemonJello
The problem was just fine the way I initially stated it. Do you seriously not understand what 'distinguishable' is there supposed to convey?

Originally posted by forkedknight
You need to consider 5 cases, and then multiply by 2

Originally posted by LemonJello
Bull. Your restatement is ambiguous and does not faithfully preserve the problem as I initially presented it, which made clear that the poles are distinguishable whereas flags of a given color are not.

What do you think, what kind of impression would you leave on yourself calling others idiots? Childish statements about you? It doesn't matter that nobody knows you (in life) here. Everyone creates the impression about him himself.

Originally posted by kbaumen
What do you think, what kind of impression would you leave on yourself calling others idiots? Childish statements about you? It doesn't matter that nobody knows you (in life) here. Everyone creates the impression about him himself.

Originally posted by LemonJello
The following is a problem I found on a national mathematics exam.

Suppose you have two distinguishable flagpoles. You also have 19 flags. Ten of these flags are identical red flags while the other 9 are identical yellow flags. Find the number of distinguishable arrangements using all of the flags in which each pole has at least one flag and no two yellow flags on either pole are adjacent.

Originally posted by eldragonfly
Look in the mirror dummkopf, your remarks are ridiculous i am only asking a question, and a simple one at that. It doesn't surprise me one bit that you and lemmonjelly are unable to answer it. Nevermind i will sort this out myself. 😠

Originally posted by SwissGambit
1375 arrangements.

Originally posted by SwissGambit
I see the error in my program now. I was thinking of the flags as one 19-digit binary number. "1"s were yellow flags. I allowed the "no adjacent 1's" rule to be broken once [because you could insert the flagpole separation at that point].

The thing I forgot to do is, if the number does NOT have any adjacent ones, then it counts for 18 hits total, beca ...[text shortened]... on for the two flagpoles can go in 18 different places. Once I added that, I got 2310 also.

Originally posted by eldragonfly
You are functionally illiterate my man. please answer the question lemmonjelly, it is a simple one. eg. what about the case for symmetry does that count as 1 combination or two? In other words the case for 2 flagpoles and 1 red and 1 yellow flag, counts as 1 combination or 2 separate combinations, ie, if you switch/exchange flags from one pole to another. 🙄

Originally posted by LemonJello
I thought it was rather strange that you would offer up the wrong answer. 😲

Originally posted by LemonJello
4*572 + 22 = 2310 (not 2299).

And that's the correct answer. Good work.