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My teacher ask a question to me.

Question is: Determine in how many ways can be rearranged the letters of the word ECEHUCDE so that the consonants and vowels are alternating.

I said it must be $\displaystyle \frac{8!}{3!\cdot 2!}=3360$. After that i realize alternating means something different.

I did 4 3 2 1 4 3 2 1 => $\displaystyle \frac{24\cdot 24}{3!\cdot 2!}=48$. He said it is wrong too.

I can't find the solution because I don't know what does alternating mean?

Thank you for your all helps :)

Good day.

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  • $\begingroup$ The solution is 4 4 3 3 2 2 1 1 + 4 4 3 3 2 2 1 1. (24*24*2)/(3!*2!) am i right? $\endgroup$ – Zafer Dec 7 '14 at 15:35
  • $\begingroup$ Yes, see my answer confirming this. $\endgroup$ – Namaste Dec 7 '14 at 16:07
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Let $c$ denote a consonant, and $v$ denote a vowel. We have four letters of each type. Consonants are C, H, D, and vowels are E, U. Alternating means an arrangement such as $\;c\,a\,c\,a\,c\,a\,c\,a\;$ or $\;a\,c\,a\,c\,a\,c\,a\,c.\;$ That is, no consonant can be next to a consonant, and no vowel can be next to a vowel.

Responding to your comment: Yes, the bold-face in your comment is correct. Good work!

In summary, we have that there are $$\dfrac{2\cdot (4!)^2}{3!2!} = 2\cdot 48 = 96$$ distinct ways to arrange the given letters such that the arrangement alternate between consonants and vowels.

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  • $\begingroup$ I can try to vote +1 but I cant :( Vote Up requires 15 reputation $\endgroup$ – Zafer Dec 7 '14 at 19:14
  • $\begingroup$ Don't worry about it. I appreciate your acceptance and good intentions! $\endgroup$ – Namaste Dec 7 '14 at 19:21
  • $\begingroup$ A ha. Now I understand all the sudden down-votes to old and unrelated questions of mine on this website. Somebody down-voted you, so you concluded it was me (being the only other user answering this question here), and now you're taking out all the frustration on me??? That's very mature! $\endgroup$ – barak manos Dec 8 '14 at 12:27
  • $\begingroup$ @barak What are you talking about? I think it is you jumping to conclusions. Your answer is fine, and helpful as an alternative approach. I saw that both you and I had been downvoted here, and I concluded it was my "stalker" in action (I get one or two downvotes almost daily on correct, unproblematic questions. That's been going on long before your entrance to MSE, so I had no reason to believe you downvoted me.) It seems both our answers were targets yesterday. $\endgroup$ – Namaste Dec 8 '14 at 12:35
  • $\begingroup$ Possibly I am jumping to conclusions, though I did have a somewhat solid ground for suspicion after seeing your comment above in conjunction with 3 or 4 down-votes that have occurred since, some of which on rather old posts of mine (such as math.stackexchange.com/q/687736/131263). In any case, if the conclusions were wrong, then I apologize. $\endgroup$ – barak manos Dec 8 '14 at 12:46
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There are $48$ different ways to put EEEU in the odd places and CCDH in the even places:

$\binom43\cdot\binom11\cdot\binom42\cdot\binom21\cdot\binom11=48$

There are $48$ different ways to put EEEU in the even places and CCDH in the odd places:

$\binom43\cdot\binom11\cdot\binom42\cdot\binom21\cdot\binom11=48$

Hence there are $96$ ways to arrange EEEUCCDH with the consonants and vowels alternating.

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