Solving probability problem with complementation approach. Problem

If it is assumed that all $\binom{52}{5}$ poker hands are equally likely, what is the probability of being dealt one pair (i.e. a,a,b,c,d where a,b,c, and d are all distinct)?

I was trying to solve this by complementation.

Let us find number of ways we can serve hand so that there is no pair. We can select five different denominations in $\binom{13}{5}$ ways. There are four cards of each domination. Thus total number of ways to serve pair-less hands is $=\binom{13}{5}\times 4^5$. To get number of ways to have at least one pair, we subtract it from total number of ways to serve a hand of five cards: $\binom{52}{5}$. So the final probability will be 
$=\frac{\binom{52}{5}-\binom{13}{5}\times 4^5}{\binom{52}{5}} = 0.4929$

The solution given was as follows:

  
*
  
*Number of ways to choose one denomination out of 13: $\binom{13}{1}$
  
  
*
  
*Number of ways to select 2 cards of selected denomination: $\binom{4}{2}$
  
*Number of ways to choose 3 denominations out of remaining 12: $\binom{12}{3}$
  
*Number of ways to select 1 card each of selected denominations: $\binom{4}{1}\times\binom{4}{1}\times\binom{4}{1}$ 
  
*Final probability $$=\frac{\binom{13}{1}\times\binom{4}{2}\times\binom{11}{3}\times\binom{4}{1}\times\binom{4}{1}\times\binom{4}{1} }{\binom{52}{5}}\approx 0.4226$$
  
  

Whats going wrong in my solution?
Update
As stated by joriki in answer, I forgot to consider other arrangements. I am listing all below:


*

*all different denominations (i.e. a,b,c,d,e): $\binom{13}{5}4^5$

*two pairs (i.e. a,a,b,b,c): $\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}$

*three of a kind (i.e. a,a,a,b,c): $\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}$

*four of a kind (i.e. a,a,a,a,b): $\binom{13}{1}\binom{12}{1}\binom{4}{1}$


So the desired probability becomes:
$$\frac{\binom{52}{5}-\binom{13}{5}4^5-\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}-\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}-\binom{13}{1}\binom{12}{1}\binom{4}{1}}{\binom{52}{5}}\approx 0.424$$
So this seems to closely match to the solution given. However, I am afraid that I might have missed something because the answer differs in fractions. So please check if I have mistaken somewhere.

You ignored the part "where $a$, $b$, $c$, and $d$ are all distinct". You counted the hands that contain at least one pair, not the ones that contain exactly one pair.
