I am trying to figure this problem out and it just makes no sense to me. The questions is as follows:
A box contains 5 Green balls and 5 Orange Balls. Five balls are taken at random without replacement. What is the probability that two green balls and 3 orange balls are selected?
Now working through the problem manually I come up with this: (5/10)(4/9)(3/8)(5/7)(4/6) = 0.039
But using hypergeometric rules the best I understand I come up with this:
((5,2)(5,0))/(10,2) = .222
and
((5,3)(5,0))/(10,3) = .0833
.222*.083 = .0185
What am I not understanding here. My confusion is in the fact that we are selecting both green and orange and not just looking for all orange or all green
Your manual calculation should have multiplied by ${5 \choose 3}$ or ${5 \choose 2}$, i.e. by $10$, and indeed it seems you have done so to get $0.39\ldots$ rather than $0.039\ldots$
If you wanted this just in terms of binomial coefficients and factorials then you could have calculated the number of arrangements of three orange balls and two green balls, and ways of choosing three out of five orange and two out of five green: $${3+2 \choose 3} \frac{\frac{5!}{3!}\frac{5!}{2!}}{\,\frac{(5+5)!}{(3+2)!}\,} = \frac{\displaystyle{5\choose3}{5\choose 2}}{\displaystyle{10 \choose 5}}.$$