Combinations of lottery balls? I'm studying for a final and practicing on old exam. I am having trouble with a question. 
"There is a lottery game that consists of two sets of numbered balls. There are 20 white balls numbered 1-20 and there are 15 red balls numbered 1-15. Four of the twenty white balls are selected and two two of the red balls are selected.
The latter ticket is white #2, #5, #14, and #17 and red #6 and #9. Find the probability that a lottery ticket matches all four correct white winning numbers and exactly one red winning number."
I know that the answer has something to do with combinations. I am getting confused because balls are numbered. It is not as simple as {(20 choose 4)*(15 choose 1)}/(35 choose 5). I know that because the balls are numbered order matters but after that I'm confused.
 A: In lottery games the order does not count 
The balls are numbered because the gamblers have to choose numbers and not because of defining some kind of order.
The total number of possibilities is
$${20\choose 4}{15\choose 2}.$$
There is only $1$ way to match the $4$ white numbers. If we want exactly one red match then we either choose the first winning number or the second winning number. In both cases our second choice is neither the first choice nor the other winning number. The total is $2\times 13=26$.
So the probability we seek is
$$\frac{26}{{20\choose 4}{15\choose 2}}.$$
A: It is not clear whether the order matters or not. I would say that only the numbers are important and not the order of them.
If the order does not matter, we have ${20\choose 4}$ ways to choose $4$ white balls and ${15\choose 2}$ ways to choose $2$ red balls. Hence, we have 
${20\choose 4}\cdot{15\choose 2}$ ways to choose $4$ white balls and $2$ red balls. There is only one way to choose $4$ correct white winning numbers, but there are $2\cdot 13$ ways to choose exactly $1$ red winning number (we have to choose $1$ winning number out of $2$ and $1$ other number out of the remaining $13$ numbers). The probability is then given by
$$
\frac{2\cdot 13}{{20\choose 4}\cdot{15\choose 2}}.
$$
If the order matters, we have $20\cdot19\cdot18\cdot17$ ways to choose $4$ white balls and $15\cdot 14$ ways to choose $2$ red balls. There is only one way to choose $4$ correct white winning numbers, but there are $2\cdot13$ ways to choose exactly $1$ red winning number since the winning number might be the number of the first white ball or the second white ball. The probability is then given by
$$
\frac{2\cdot13}{(20\cdot19\cdot18\cdot17)\cdot(15\cdot14)}.
$$
A: probability $4$ winning white: $1\over{20\choose 4}$, $2$ winning red: $1\over{15\choose 2}$
probability $1$ winning red(choose which winning ball and which non-winning ball): ${{2\choose 1}{13\choose 1}}\over{15\choose 2}$
prob.  $4$ winning white and $1$ winning red: ${1\over{20\choose 4}}\times {{{2\choose 1}{13\choose 1}}\over{15\choose 2}}={{2\times 13}\over {{20\choose 4}{15\choose 2}}}$
