Select the last 3 child elements - css

Question

I know you can select the last child with :last-child or a certain child with :nth-child (if you know their position/number).
What I need is to select the last 3 children without knowing how many child elements there could be.
I know there is something that's called :nth-last-child but I cant really understand how it is working.
For this:
<div id="something">
<a href="images/x.jpg" ><a/>
<a href="images/x.jpg" ><a/>
<!-- More elements here -->
<!-- ..... -->
<!-- I need to select these: -->
<a href="images/x.jpg" ><a/>
<a href="images/x.jpg" ><a/>
<a href="images/x.jpg" ><a/>
</div>
I need something like this:
#something a:last-child{
/* only now it selects the last <a> and the 2 <a>'s that come before*/
}

Answer 1

You can read more here about nth-last child, but this should basically do the trick of selecting the last 3 children with just CSS
#something a:nth-last-child(-n+3) {
/*declarations*/
}
fiddle demonstration from Fabrício Matté
This will only select those rows returning a positive number for out N expression (-n+3), and since we are using nth-last-child, it's counting from last to first,
so first rows from bottom gives,
f(n) = -n+3
f(1) = -1+3 = 2 <- first row from the bottom
f(2) = -2+3 = 1 <- second row from the bottom
f(3) = -3+3 = 0 <- third row from the bottom
everything else will return a negative number

Answer 2

This is possible with CSS3 selectors and formulas:
.something a:nth-last-child(-n+3) { ... }
You could also try using jQuery (example) or adding a specific class to the last three elements in your server-side code (it does not require JavaScript to be enabled in browsers and also works on older browsers that do not support CSS3).

Answer 3

The accepted answer has the correct formula, but the explanation is wrong.
So the correct CSS is (same as currently accepted answer):
#something a:nth-last-child(-n+3) {
/*declarations*/
}
But here is the correct explanation of the math:
f(n) = -n+3
f(0) = -0+3 = 3 <- 3rd row from the bottom
f(1) = -1+3 = 2 <- 2nd row from the bottom
f(2) = -2+3 = 1 <- 1st row from the bottom
f(3) = -3+3 = 0 <- nothing
f(3) = -4+3 = -1 <- nothing
etc...