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In how many ways can letters in a word CALCULUS be rearranged so that no two identical letters are adjacent?

I’ve been thinking of Inclusion-Exclusion principle. Is there any different way to solve this task?

N. F. Taussig
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retne
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    Inclusion-Exclusion should work fine. You have three letters which appear twice, so you will not get that many terms in the final expansion for the rearrangements. – EuxhenH Jan 07 '19 at 10:28
  • I know but I wonder if there’s any other way to do that not using sophisticated methods – retne Jan 07 '19 at 11:26
  • Not using sophisticated methods would force you to do tedious casework. – N. F. Taussig Jan 07 '19 at 16:02
  • What would be the final answer then – retne Jan 08 '19 at 08:52
  • If you want additional feedback, then you should [edit] your post to show what you have attempted and explain where you are stuck. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Jan 08 '19 at 10:56
  • Thank you! I will edit within a few hours – retne Jan 08 '19 at 11:22
  • Have you made any progress? – N. F. Taussig Jan 11 '19 at 09:30
  • I counted all the permutations: 7! but I’m not sure of the usage of inclusion-exclusion principle. There exists formula : sum from p=1 to n (-1)^(p-1) times n!/p! but don’t know how to apply it. – retne Jan 11 '19 at 13:48
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    There are eight letters in CALCULUS, of which two are Cs, two are Ls, two are Us, one is a A, and one is an S. If we did not have the restriction that no two adjacent letters may be the same, we could choose two of the eight positions for the Cs, two of the remaining six positions for the Ls, two of the remaining four positions for the Us, one of the remaining two positions for the A, and then fill the remaining position with the S. We can do this is in $$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{1} = \frac{8!}{2!2!2!1!1!}$$ distinguishable ways. – N. F. Taussig Jan 12 '19 at 11:49
  • My answer to this question may help you understand how to apply the Inclusion-Exclusion Principle to the problem you asked. – N. F. Taussig Jan 12 '19 at 11:50
  • Any further progress? – N. F. Taussig Jan 14 '19 at 12:29
  • I’m pretty busy right now, but I’ll try to get that next week. – retne Jan 15 '19 at 15:17

1 Answers1

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The answer is $2220$

CALCULUS automaton here is an automaton that generates CALCULUS-like words. To cover all possibilities, the S state must and suffice to be reached 5 to 8 times.

For each time, the generating function is

$$Z = (c+a+l+u+s - c^2 - l^2 - u^2)$$

where $+$ stands for inclusion, and $-$ for exclusion.

Now we take the coefficient of $c^2al^2u^2s$ in $Z^5+Z^6+Z^7+Z^8$ which is $2220$.

The explanation is that each occurence of a double letter in each word is generated twice, once with + and once with -. After summing, the terms containing double consecutive letters vanish.

There is no simpler method. If there were, the generating function would be simpler but that is not the case since I checked them all against the problems here.

Boyku
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