Is there an easy way in .NET to get "st", "nd", "rd" and "th" endings for numbers?

It's a function which is a lot simpler than you think. Though there might be a .NET function already in existence for this, the following function (written in PHP) does the job. It shouldn't be too hard to port it over.

function ordinal($num) {
    $ones = $num % 10;
    $tens = floor($num / 10) % 10;
    if ($tens == 1) {
        $suff = "th";
    } else {
        switch ($ones) {
            case 1 : $suff = "st"; break;
            case 2 : $suff = "nd"; break;
            case 3 : $suff = "rd"; break;
            default : $suff = "th";
        }
    }
    return $num . $suff;
}

Simple, clean, quick

    private static string GetOrdinalSuffix(int num)
    {
        string number = num.ToString();
        if (number.EndsWith("11")) return "th";
        if (number.EndsWith("12")) return "th";
        if (number.EndsWith("13")) return "th";
        if (number.EndsWith("1")) return "st";
        if (number.EndsWith("2")) return "nd";
        if (number.EndsWith("3")) return "rd";
        return "th";
    }

Or better yet, as an extension method

public static class IntegerExtensions
{
    public static string DisplayWithSuffix(this int num)
    {
        string number = num.ToString();
        if (number.EndsWith("11")) return number + "th";
        if (number.EndsWith("12")) return number + "th";
        if (number.EndsWith("13")) return number + "th";
        if (number.EndsWith("1")) return number + "st";
        if (number.EndsWith("2")) return number + "nd";
        if (number.EndsWith("3")) return number + "rd";
        return number + "th";
    }
}

Now you can just call

int a = 1;
a.DisplayWithSuffix(); 

or even as direct as

1.DisplayWithSuffix();

No, there is no inbuilt capability in the .NET Base Class Library.