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My attempt:

Consider the complement where all the consonants are considered as one unit: $BBBLLCS$

Step 1: Arrange the consonants within the unit: $\displaystyle\frac{7!}{3!2!}$

Step 2: Arrange the vowels: $\displaystyle\frac{6!}{2!2!}$

The total number of arrangements of the $13$ letters with no restriction is $\displaystyle\frac{13!}{3!(2!)^3}$

So the total number of arrangments where no two consonants are next to each other is

$$\displaystyle\frac{13!}{3!(2!)^3}-\left[\displaystyle\frac{7!}{3!2!}\cdot \displaystyle\frac{6!}{2!2!}\right]$$

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    $\begingroup$ I don't think so. Can you write down an example of an arrangement with no two consonants next to one another? There really aren't a lot of recipes for these counting problems. You have to approach each on its own. Sure you can often apply techniques from other problems you've solved, but you can't apply them blindly. $\endgroup$ – saulspatz Oct 15 '18 at 22:22
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    $\begingroup$ @rover2 Not if you place a vowel between each pair of consonants. $\endgroup$ – N. F. Taussig Oct 15 '18 at 22:28
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    $\begingroup$ You have to start and end with a consonant. $\endgroup$ – saulspatz Oct 15 '18 at 22:32
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    $\begingroup$ okay im following...so it has to be CVCVCVCVCVCVC...there are $\frac{7!}{3!2!}$ ways to arrange the consonants and $\frac{6!}{2!2!}$ ways to arrange the vowels...but how do i go about taking care of this restriction $\endgroup$ – rover2 Oct 15 '18 at 22:37
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    $\begingroup$ You just did. For each of the $\frac{7!}{3!2!}$ ways of arranging the consonants in the positions reserved for consonants, you can arrange the vowels in the positions reserved for vowels in $\frac{6!}{2!2!}$ ways. $\endgroup$ – N. F. Taussig Oct 15 '18 at 22:51
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There are 13 letters and 7 consonants. Since no two consonants can be together, this means there is a "space" between each consonant, dividing the possibilities into $C_ C_ C_ C_ C_ C_ C$ where C means Consonant. There is only one way for the general structure, but how do we choose which consonants for each spot? Since there are 3 B's, 2 L's, and one each of C and S, what does that give you? Hint: $\binom{7}{3}$ ways to place the B's, leaving 4 consonants; $\binom{4}{2}$ ways to place the L's, and so on.

I agree with you on arranging the vowels; since there is no other restriction, there are 6! ways to place vowels, but this double counts the U's (divide by $2$) and double counts the I's (divide by another 2).

The total is the product of the two.

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    $\begingroup$ as in the comments above, this gives us $\frac{7!}{3!2!}\cdot \frac{6!}{2!2!}$ $\endgroup$ – rover2 Oct 15 '18 at 23:19

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