How to obtain the number of RNA chains after a selective fragmentation? The problem is as follows:

A molecular scissor is an enzyme which breaks down RNA chain. At a
  laboratory a technician uses a G-enzyme into the chain CCGGUCCGAAAG and
  it gets broken in the sites where it detects a G base, obtaining a
  fragment CCG-G-UCCG-AAAG. If a U-enzyme and a C-enzyme were used to a
  certain RNA chain, the following fragments were obtained:
  C-C-GGU-C-C-GAAAG. Find from how many different chains of RNA could
  come such fragments?.

The existing alternatives given are:
$\begin{array}{ll}
1.&5\\
2.&30\\
3.&720\\
4.&120
\end{array}$
From the looks of the title it seems that is has to do with combinatorics but I'm totally lost with this one. Can somebody help me?. My best guess at this point is that because a specific sequence is required then the order it matters hence it should be a permutation.
Because there appears only four bases CGUA and the chain has 12 bases in total and the fragments are two for U and C, then I thought it could be as this:
$P=\frac{12!}{\left(12-4\right)}=\frac{12\times 11 \times 10 \times 9 \times 8!}{8!}=11880$
But the number I obtained is too big and it doesn't get any closer to the alternatives. Therefore I need help with this one. Can somebody put me in the right direction?.
 A: If I understand the problem correctly, we know that we got four C fragments, one GGU fragment, and one GAAAG fragment, but we don't know the order they came in.  We do know that a U-enzyme and a C-enzyme were used to split the chain.  Now, since there was no G enzyme used for splitting, the GAAAG fragment must come last, leaving us with four C's and a GGU.  These can come in any order, so there are five possibilities.  You can see this by noting that there are five places to put the GGU among the C's (including before the first and after the last) or by using the formula $${5!\over4!}=5.$$  Let me just note that the second answer, $30,$ is a trap for for those who say that the answer is $${6!\over4!}=30$$ without realizing that GAAAG must come last.
EDIT
To count permutations with repeated elements, you divide by the factorial of the number of repeats.  In this case, we have four $C$'s.  Call them $C_1, C_2, C_3, C_4.$  When we treat them as distinct, just four of the six fragments, we get $6!$ arrangements.  But the C's are really the same, so $C_1C_2C_3C_4$ is no different from $C_4C_3C_2C_1$, for example.  Therefore, we counted every arrangement $4!$ times, and we have to divide by $4!$.  I we had some other fragment repeated $3$ times, for example, we'd also have to divide by $3!$ for the same reason.         
