Combinatorics/Probability - Multiple Groups Example Problem 
Joe, an avid and properly licensed sportsman, is in his hunting blind when he locates 20 Canada geese, 25 Mallard ducks, 40 Bald Eagles, 10 Whopping Cranes, and 5 Flamingos.  Joe randomly selects six birds to target.  What is the probability that at least one of each species is targeted?

So we have 5 birds and we must choose 6:
20 g 
25 d 
40 e 
10 c 
5 f 
So we have six slots and one slot will be a wild card.  So we must calculate the outcomes for the following:  {g,d,e,c,f,*}, where *=[g or d or e or c or f]
$Pr=\cfrac{\{g,d,e,c,f, *\}}{\binom{100}{6}}$
My approach to calculating {g,d,e,c,f,*} is to manually enumerate the 5 cases and sum them together:


*

*ggdecf = $\binom{20}{2}*25 * 40 * 10 * 5$

*gddecf = $20*\binom{25}{2} * 40 * 10 * 5$

*gdeecf = $20 * 25*\binom{40}{2} * 10 * 5$

*gdeccf ...

*gdecff ...
I verified that this gives me the correct answer of $0.0398472$.  
I just wanted to know if there's a more efficient way of doing this then what I am doing.  Am I approaching the question properly? 
I thought about composing a vector (g,d,e,c,f) where each coordinate is like an "urn" and we will be placing 6 balls (the bullets) inside them.  So we could have (2,1,1,1,1) which is 2 geese and the rest of the birds are one.  If we wanted to find all the combinations of 1 pair, we could use the multinomial theorem:
$\binom{5}{2,1,1,1,1}=60$.  But we still have to calculate the probability of each pair and multiply it by this right?  Can you do it this way?  
I know usually when a question asks you at least you should subtract the cases so save you work.  Can you do that in this case?
Thank you.

See my answer here to a similar problem involving 3 choices. I am sure you can extend it.
