Question 912042
<pre>
We choose the 4 in 9C4 ways.  That's {{{(9*8*7*6)/(4*3*2*1)}}}
We choose the 3 in 5C3 ways.  That's {{{(5*4*3)/(3*2*1)}}}
We choose the 2 in 2C2 ways.  That's {{{(2*1)/(2*1)}}}

Multiplying all those together, that's

{{{(9*8*7*6*5*4*3*2*1)/((4*3*2*1)*(3*2*1)*(2*1))}}}

That's {{{9!/(4!3!2!)}}}

That's not listed, but I'm guessing choice c was supposed to be 9!/(4!3!2!).

Edwin</pre>