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$P(E) = \dfrac{\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...}{N!}$

How can I force the red onto the next line of the numerator?

Moreover, how can I add an empty line (ie double space) the numerator?

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\\ works:

$$P(E) = \dfrac{\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) }\\\color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...}{N!}$$

$$P(E) = \dfrac{\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) }\\\color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...}{N!}$$

Double line (using the non-breaking space character ~):

$$P(E) = \dfrac{\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) }\\~\\\color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...}{N!}$$

$$P(E) = \dfrac{\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) }\\~\\\color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...}{N!}$$

Or use the align environment:

$$P(E) = \dfrac{\begin{align}&\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \\ &\color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...\end{align}}{N!}$$

$$P(E) = \dfrac{\begin{align}&\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \\ &\color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...\end{align}}{N!}$$

$$P(E) = \dfrac{\begin{align}&\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \\ \\&\color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...\end{align}}{N!}$$

$$P(E) = \dfrac{\begin{align}&\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \\ \\&\color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...\end{align}}{N!}$$

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    $\begingroup$ NB. You should really use \ldots in place of .... $\endgroup$ – Lord_Farin Nov 20 '13 at 14:13
  • $\begingroup$ You might want to include a multline version as well. (IMHO it looks better than the align.) $\endgroup$ – user642796 Nov 20 '13 at 14:16
  • $\begingroup$ @Lord_Farin: Many thanks. Upvoted. Can a new empty line (For double spacing) be forced with // or something other than \align? Please answer in your answer; I'd like to consolidate all content. $\endgroup$ – Greek - Area 51 Proposal Nov 20 '13 at 14:28
  • $\begingroup$ @ArthurFischer I can't get multline to work -- it looks hideously misaligned to me. $\endgroup$ – Lord_Farin Nov 20 '13 at 15:03
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    $\begingroup$ @Lord_Farin: Maybe we have different aesthetic tastes in this matter. I for one think that the "misalignment" is a good indicator that something has spanned more than one line. $\endgroup$ – user642796 Nov 20 '13 at 15:14
  • $\begingroup$ @ArthurFischer To me, the expression isn't even properly lined up above the bar -- it's shifted to the right by some length. But indeed, my taste here is to have both lines start without indentation. In an eqnarray-type environment, I agree that indentation is helpful. $\endgroup$ – Lord_Farin Nov 20 '13 at 15:26
  • $\begingroup$ @Lord_Farin: I think it has something to do with how frac works. In neither align nor multline is the formula centred above the bar. It is perfectly "right-aligned" but not perfectly "left-aligned". (Or at least this is how I am currently seeing it.) $\endgroup$ – user642796 Nov 20 '13 at 15:41
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In addition to Lord_Farin's options, I'll include one-and-a-half using multline (which he apparently does not favour).

$$P(E) = \dfrac{\begin{multline}\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \\ \color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...\end{multline}}{N!}$$

yields

$$P(E) = \dfrac{\begin{multline}\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \\ \color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...\end{multline}}{N!}$$

Introducing a second \\ at the obvious place results in

$$P(E) = \dfrac{\begin{multline}\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \\ \\ \color{red}{\times N} - k \text{ choices for C(N - k) }\times N - k - 1 \text{ choices for C(N - k - 1)} \times ...\end{multline}}{N!}$$

[It appears that some users will see some hideous monstrosity. Apologies.]

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    $\begingroup$ Here's how it looks to me: picture. Do you still blame me for not including it? $\endgroup$ – Lord_Farin Nov 20 '13 at 15:43
  • $\begingroup$ @Lord_Farin: Yikes, that's horrible! It actually looks quite pleasant to me. (And your examples don't look that bad either.) $\endgroup$ – user642796 Nov 20 '13 at 15:51
  • $\begingroup$ +1 on the basis of your screenshot. And thanks for exposing me to multline! $\endgroup$ – Lord_Farin Nov 20 '13 at 16:16
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    $\begingroup$ multline is intended for top-level use only (LaTeX prevents its use in other ways, but MathJax is not so picky). It creates elements that try to be 100% wide, and that can cause it to overflow the fraction. It appears that browsers differ on how they handle that. In any case, this is not really valid LaTeX, and I wouldn't recommend it. $\endgroup$ – Davide Cervone Dec 3 '13 at 21:17
  • $\begingroup$ I've got the same result as Lord_Farin on Google Chrome. $\endgroup$ – Pacerier Sep 12 '14 at 13:05

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