Originally posted by kbaumenLook 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. ðŸ˜
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 LemonJello1375 arrangements.
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 eldragonflyWell, it was answered, in case you didn't notice.
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. ðŸ˜
Originally posted by SwissGambitI 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].
1375 arrangements.
The thing I forgot to do is, if the number does NOT have any adjacent ones, then it counts for 18 hits total, because the separation for the two flagpoles can go in 18 different places. Once I added that, I got 2310 also.
Originally posted by SwissGambitI thought it was rather strange that you would offer up the wrong answer. 😲
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.
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Originally posted by eldragonflyTo answer your question, the two poles are distinguishable. So if we consider your scenario in which a red flag is on pole 1 and a yellow flag is on pole 2, that is a separate arrangement from a scenario in which a yellow flag is on pole 1 and a red flag is on pole 2. Hope that helps, sweetie pie.
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. 🙄