- 27 May '08 22:20The 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. - 30 May '08 03:29

Why would I want to do something like that?*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. - 30 May '08 17:08

combinations or permutations, clean this up if you would..*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. - 30 May '08 18:32 / 2 edits

"distinguishable" implies permutations, but a permutation using different but identical members is ignored.*Originally posted by eldragonfly***combinations or permutations, clean this up if you would..**

for example

XXO is different from XOX, but swapping the Xs results in the same permutation of XXO, so you only count it once. - 30 May '08 18:41this problem is too time intensive for me because you have to account for all possibilities of different numbers of each type of flag used

the problem would be difficult enough if you had to use all 19 flags, but allowing different numbers creates a very very large number of possibilities. - 30 May '08 20:49 / 2 edits

The problem stipulates that you are to "find the number of distinguishable arrangements*Originally posted by forkedknight***this problem is too time intensive for me because you have to account for all possibilities of different numbers of each type of flag used**

the problem would be difficult enough if you had to use all 19 flags, but allowing different numbers creates a very very large number of possibilities.**using all of the flags**in which each pole has at least one flag and no two yellow flags on either pole are adjacent." Among other things, that is intended to mean that you are limited to arrangements that use all 19 flags, which greatly constrains the problem.

You seem to be considering a different problem, which I would agree with you is a more time intensive one. - 30 May '08 21:27 / 1 edit

this is silly lemmonjelly.*Originally posted by LemonJello*

You have two flagpoles.

You also have 19 flags.

There are 10 red flags and there are 9 yellow flags.

Find the number of separate combinations using all of the flags in which each pole has at least one flag and no two yellow flags on either pole are adjacent. - 30 May '08 21:52 / 1 edit

Well, given our interactions in the past, you must know by now I reserve precious little respect for your mathematical endowments. I place your thoughts and input on such matters somewhat below those of the retarded.*Originally posted by eldragonfly***this is silly lemmonjelly.** - 30 May '08 21:52You need to consider 5 cases, and then multiply by 2

Number of yellow flags on a pole:

9:

one red flag on the other pole, and 8 flags in between the yellows, so there are 11 places for the last red flag to go

11 * 2 = 22

8:

one yellow flag on the other pole, and 7 flags in between the Y's

therefore 3 red flags and 11 places to put them, so

C(11,3) + 2 * C(11,2) + 11 = (165+ 110 +11) * 2 = 572

7:

once again, 7 R's in between the Y's and 3 remianing R's to place

= 572

6:

same deal

= 572

5:

same again

= 572

Therefore 4 * 572 + 22 = 2299 ways - 30 May '08 21:54

4*572 + 22 = 2310 (not 2299).*Originally posted by forkedknight***You need to consider 5 cases, and then multiply by 2**

Number of yellow flags on a pole:

9:

one red flag on the other pole, and 8 flags in between the yellows, so there are 11 places for the last red flag to go

11 * 2 = 22

8:

one yellow flag on the other pole, and 7 flags in between the Y's

therefore 3 red flags and 11 places to put them, so ...[text shortened]... 72

6:

same deal

= 572

5:

same again

= 572

Therefore 4 * 572 + 22 = 2299 ways

And that's the correct answer. Good work. - 31 May '08 00:01 / 1 edit

i only restated your problem in clear language, you don't know me at all ; that makes your childish statements and namby pampy conclusions a bit strange my man.*Originally posted by LemonJello***Well, given our interactions in the past, you must know by now I reserve precious little respect for your mathematical endowments. I place your thoughts and input on such matters somewhat below those of the retarded.** - 31 May '08 00:41You have two flagpoles.

You also have 19 flags.

There are 10 red flags and there are 9 yellow flags.

Find the number of separate combinations using all of the flags in which each pole has at least one flag and no two yellow flags on either pole are adjacent.

this problem is still poorly worded lemmonjelly, 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. - 31 May '08 01:01 / 2 edits

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 eldragonfly***You have two flagpoles.**

You also have 19 flags.

There are 10 red flags and there are 9 yellow flags.

Find the number of separate combinations using all of the flags in which each pole has at least one flag and no two yellow flags on either pole are adjacent.

this problem is still poorly worded lemmonjelly, eg. what about the case for symmetry do ...[text shortened]... mbination or 2 separate combinations, ie, if you switch/exchange flags from one pole to another. - 31 May '08 01:06 / 2 edits

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.*Originally posted by eldragonfly***i only restated your problem in clear language,** - 31 May '08 16:06You 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.