What does alternating mean?

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.

• 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? Commented Dec 7, 2014 at 15:35
• Yes, see my answer confirming this. Commented Dec 7, 2014 at 16:07

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.

• I can try to vote +1 but I cant :( Vote Up requires 15 reputation Commented Dec 7, 2014 at 19:14
• Don't worry about it. I appreciate your acceptance and good intentions! Commented Dec 7, 2014 at 19:21
• 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! Commented Dec 8, 2014 at 12:27
• @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. Commented Dec 8, 2014 at 12:35
• 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. Commented Dec 8, 2014 at 12:46

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.