[SOLVED] \frac with \choose

• Jan 20th 2010, 05:13 AM
courteous
[SOLVED] \frac with \choose
How can one present fraction with combinations above and below?

For example, \frac{ 4\choose 4 48\choose 9 }{ 52\choose 13 } gives this: $\frac{ 4\choose 4 48\choose 9 }{ 52\choose 13 }$

I've already searched for "choose" in "Latex Help Forum".
• Jan 20th 2010, 07:38 AM
Plato
Quote:

Originally Posted by courteous
How can one present fraction with combinations above and below?

For example, \frac{ 4\choose 4 48\choose 9 }{ 52\choose 13 } gives this: $\frac{ {4 \choose 4}{48 \choose 9} }{ 52 \choose 13 }$

I've already searched for "choose" in "Latex Help Forum".

I would use \binom{}{}
$$\frac{\displaystyle\binom{4}{4}\binom{48}{9}}{\dis playstyle\binom{52}{13}}$$

$\frac{\displaystyle\binom{4}{4}\binom{48}{9}}{\dis playstyle\binom{52}{13}}$
• Jan 20th 2010, 09:27 AM
Soroban
Hello, courteous!

Quote:

How can one present fraction with combinations above and below?

For example, \frac{ 4\choose 4 48\choose 9 }{ 52\choose 13 } gives this: $\frac{ 4\choose 4 48\choose 9 }{ 52\choose 13 }$

The "choose" function must be enclosed in braces: .{6 \choose 2}

The code looks like this:

\frac{ {4\choose4} {48\choose9} } { {52\choose13} }

. . . $\frac{{4\choose4}{48\choose9}}{{52\choose13}}$

As Plato pointed out, it can be enlarged
. . by inserting \displaystyle in the numerator and denominator.

\frac{\displaystyle {4\choose4} {48\choose9} } {\displaystyle {52\choose13} }

. . . $\frac{\displaystyle{4\choose4}{48\choose9}}{\displ aystyle{52\choose13}}$

• Jan 21st 2010, 01:49 AM
courteous
$\frac{{4\choose 4}{48\choose 9}}{{52\choose 13}}$
It works!
Though I'm pretty sure I tried curly braces around \choose before. (Thinking)

Thank you both. (Happy)