Forming a sequence from a given set 28 random draws are made from the set {1,2,3,4,5,6,7,8,9,A,B,C,D,J,K,L,U,X,Y,Z} containing 20 elements. In each draw, one element from the set is drawn with replacement.
What is the probability that the sequence CUBAJULY1987 occurs in that order in the chosen sequence?
The answer given is $$\frac{17 \times 20^{16} - 15 \times 20^{4}}{20^{28}}$$
I'm not sure how to approach the question. This is what I have so far. 
1)The total number of ways to draw 20 elements in a particular draw is $20!$.
2) In the sequence "CUBAJULY1987", the chance that every single element is drawn correctly is $P_{12}^{20}$?
 A: Since we are drawing $28$ times, it is clear that the sampling is done with replacement. There are $20^{28}$ equally likely strings of length $28$ over an alphabet of size $20$.  
We now count the favourables. Because our desired string CUBAJULY1987 can occur twice, for the counting we will use a simple form of Inclusion/Exclusion. 
Our desired string can occur with the initial C starting at position $1$, $2$, $3$, and so on up to position $17$.  So there are $17$ places where it can start.  Then the rest of the $16$ positions can be filled in $20^{16}$ ways, giving a total of $17\cdot 20^{16}$.  
However, this double-counts the sequences in which our string occurs twice. If that happens, that leaves $4$ "free" positions. These positions can occur before the first occurrence of our string, between the two occurrences, or after the second occurrence, or in some combination thereof.  The number of choices for where the junk symbols occur is the number of solutions of $x_1+x_2+x_3=4$ in non-negative integers. This is (Stars and Bars, or simple enumeration of cases) $\binom{6}{2}$. Once we have chosen the locations of the positions of the junk letters, they can be filled in $20^4$ ways for a total of
$\binom{6}{2}\cdot 20^4$ ways. 
It follows that the number of favourables is
$$17\cdot 20^{16}-\binom{6}{2}\cdot 20^4.$$
For the probability, divide by $20^{28}$. 
Remark: Please note there is a typo in the numerator of the answer as given in the post. It should read $15\times 20^4$, not $15\times 20^{14}$. 
