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    <p>I think the total number of variants is 466. </p> <p>One can compute the number as follows: </p> <p>If we assume that a E-E link is marked as a Y then for example the total number of arrangement s in which only two E's are linked out of 12 is equal to the number of arrangements with repetitions of 2 items when the first is taken to be repeated 10 times and the second is taken to be repeated only one time. Basically this amounts to the following list: </p> <pre><code>Y E E E E E E E E E E E Y E E E E E E E E E E E Y E E E E E E E E .. E E E E E E E E E E Y </code></pre> <p>This is basically the same as computing multinomial(10, 1) which is 11 ( <a href="http://www.wolframalpha.com/input/?i=multinomial%2810%2C+1%29" rel="nofollow">http://www.wolframalpha.com/input/?i=multinomial%2810%2C+1%29</a> )</p> <p>the total number is the following sum: </p> <pre><code>multinomial(12) + // there is no E-E link at all multinomial(12 - 2, 1) + // only one E-E link multinomial(12 - 4, 2) + // two E-E links ... multinomial(12 - 12, 6) // 6 E-E links </code></pre> <p>this is 233 ( <a href="http://www.wolframalpha.com/input/?i=multinomial%2812%29+%2B+multinomial%2810%2C+1%29+%2B+multinomial%288%2C+2%29+%2B+multinomial%286%2C+3%29+%2B+multinomial%284%2C+4%29+%2B+multinomial%282%2C+5%29+%2B+multinomial%286%29" rel="nofollow">http://www.wolframalpha.com/input/?i=multinomial%2812%29+%2B+multinomial%2810%2C+1%29+%2B+multinomial%288%2C+2%29+%2B+multinomial%286%2C+3%29+%2B+multinomial%284%2C+4%29+%2B+multinomial%282%2C+5%29+%2B+multinomial%286%29</a> )</p> <p>normally i would just multiply this by 2 to account for the end link possibility but this can actually create a chain of three E-E-E in some cases which will break your problem. </p> <p>On the other side there are available algorithms to generate all the multinomial combinations for some parameters and since all of them are like 200 it's easy to generate all and just check which ones can be extended with the circular link at then end.</p>
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