Math Is Fun Forum

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,876

Re: counting

Another way of saying it:

OE, EA & AO are invalid arrangements, each numbering 5! = 120.

If OE is immediately preceded by A (ie, AOE) or immediately followed by A (ie, OEA) - it can't be both - that reduces AO's or EA's total (respectively) by 4!, viz: 5!-4! = 120-24 = 96.

Similarly, this scenario for OE also applies to EA & AO (which therefore also number 96), and so we have 3*96 = 288 invalid arrangements.

Total number of valid arrangements = 720 possibles-288 invalids = 6!-3(5!-4!) = 432.

Last edited by phrontister (2017-02-13 10:39:49)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,876

Re: counting

Hi thickhead;

After rereading your post #20 I found some more errors than the one I mentioned in my post #23.

Your post:

N(AE)=no.of arrangements where A and E come together as AE=5!=120
Similarly N(EO=N(OA)=120
N(OAE)=4!=N(AEO)=24
N(AE-OAE-AEO)=Number of arrangements where AE is neither immediately preceded by O nor immediately followed by O=120-24-24=72
N(OA-OAE)=120-24=96
N(EO-AEO)=120-24=96
So the number of invalid arrangements=96+96+72+24+24=312
No of valid arrangements=720-312=408

I see that in your method all lines except the last are about finding the total number of invalid arrangements, and to use that result in the last line to obtain the number of valid arrangements.

The errors I found:
1. All 2-letter orders are wrong: eg, in your first line, E follows A (as AE), which is a valid arrangement instead of the invalid one (EA) it should be.
    (a) This also affects all 3-letter arrangements.
2. In line 3 you only have two 3-letter arrangements instead of three, and both are valid arrangements instead of invalid.
3. Your "neither......nor" in line 4 should be "either......or" (and that would also apply to lines 5 and 6).
4. Lines 5 and 6 should each have two 3-letter arrangements deducted instead of one.

If I'm right, then I think your method should read like this:

N(EA)=no. of arrangements where A and E come together as EA=5!=120
Similarly N(OE)=N(AO)=120
N(OEA)=4!=N(EAO)=N(AOE)=24
N(EA-OEA-EAO)=Number of arrangements where EA is either immediately preceded by O or immediately followed by O=120-24-24=72
N(AO-EAO-AOE)=120-24-24=72
N(OE-AOE-OEA)=120-24-24=72
So the number of invalid arrangements=72+72+72+24+24+24=288
No of valid arrangements=720-288=432

This corrected method is basically what I used to arrive at my formula in post #22.

Last edited by phrontister (2017-02-10 12:31:24)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson