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.
$\endgroup$ 22 Answers
$\begingroup$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.
$\endgroup$ 8 $\begingroup$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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