Discrete probobility distributions

[steller]

Okay, so my book is horrible and it doesnt explain things very well and i cant seem to find my answer anywhere online.

I wont write out the example in the book since i think its irrelevant. I am struggling with the notation and how they get the answer.

$\displaystyle f(0) = P(X=0) = \frac{\left[\begin{array}{cc}3 \\0 \\\end{array}\right] \left[\begin{array}{cc}17 \\2 \\\end{array}\right]}{\left[\begin{array}{cc}20 \\2 \\\end{array}\right]} = \frac{68}{95}$

I m very confused on how they got the answer.
I think the (3 0 ) = 0?

Can someone please explain and
work this out because i am completely lost and i am sure its easy.
I am wiling to look at a webpage if i knew what to look up.

Thank you

 

[soroban]

Hello, steller!

You are expected to know that $\displaystyle 0! = 1.$


Then we have:

. . $\displaystyle \displaystyle{3\choose0} \:=\:\frac{3!}{0!\,3!} \:=\:1$

. . $\displaystyle \displaystyle {17\choose2} \:=\:\frac{17!}{2!\,15!} \:=\:\frac{17\cdot16}{2\cdot 1} \:=\:136$

. . $\displaystyle \displaystyle{20\choose2} \:=\:\frac{20!}{2!\,18!} \:=\:\frac{20\cdot19}{2\cdot1} \:=\:190$

Therefore: .$\displaystyle \displaystyle \frac{{3\choose0}{17\choose2}}{{20\choose2}} \:=\: \frac{1\cdot136}{190} \:=\: \frac{68}{95}$

 

[steller]

Thank you so much!!!! This was excellent. I did take calculus 3 over a year ago.