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I attempted to solve this by counting the number of combinations that have 2, 3, and 4 Es consecutive, and then subtracting that count from the total. That turned out way, way more complicated. Is there a better way to do this?

I know the total different strings is $11!/(2!3!4!)$, but can't seem to go from there.

Bob
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  • No matter where you place the $E$'s there are always $\frac{7!}{2!3!}$ strings with the $E$'s in that position. How many ways to place the $E$'s? – Asinomás Feb 23 '15 at 01:24
  • See here for the problem as to how we should place the E's http://math.stackexchange.com/a/1030161/33907 – Asinomás Feb 23 '15 at 01:25

2 Answers2

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You have seven non-Es which can be temporarily visualized as * * * * * * *. You may choose four in between positions (including the ends) for the Es. This gives you ${8\choose 4}$ ways to position the Es among the non-Es. Then you can decide which of the other letters go in the non-E slots. This can be done in ${7\choose 3}{4\choose 2}\cdot 2$ ways (select $3$ postions for the Cs, two of the remaining positions for the Os, and then arrange the H and N in the remaining two positions).

So overall there are ${8\choose 4}{7\choose 3}{4\choose 2}\cdot 2$ strings.

paw88789
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  • Thank you very much. Never occurred to me to try the divider method. – Bob Feb 23 '15 at 01:26
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    Nice, I would just like to comment it is slightly more straithforward to calculate $\binom{7}{1,1,2,3}=\frac{7!}{1!1!2!3!}$ than $\binom{7}{3}\binom{4}{2}$. Although only slightly :) – Asinomás Feb 23 '15 at 01:28
  • @bob you can accept it though. And that would give you more rep. – Asinomás Feb 23 '15 at 01:29
  • Hi Bob, as the question asker, I think you can still select an answer to accept, even if you can't do upvotes yet. – paw88789 Feb 23 '15 at 01:29
  • Thanks guys, accepted the answer.

    @TheEmperorofIceCream: If we do the 7!/(1!1!2!3!) method, doesn't that get rid of the x2 at the end?

     – Bob Feb 23 '15 at 01:34
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    The $2$ at the end was added artificially. But the equality holds as $\binom{7}{3}\binom{4}{2}\cdot 2=\frac{7!}{4!3!}\frac{4!}{2!\cdot 2!}2=\frac{7!}{3!2!}=\frac{7!}{1!1!2!3!}$ as was claimed. :) – Asinomás Feb 23 '15 at 01:37
  • The expression $\binom{7}{1,1,2,3}$ is what is sometimes called a multinomial coefficient. They are somewhat useful. Although they don't really do anything you cant figure out yourself using binomial coefficients. (They do save calculation time though) – Asinomás Feb 23 '15 at 01:39
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Start with this: how many ways can you place the letters XXXXXXXEEEE such that no two Es are consecutive? This is found thus: *EX*EX*EX*E* -- you have four Xs left to place among the five *s. Once you've found that number, you can simply permute the H, N, Os, and Cs among the Xs.

Dan Uznanski
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