# 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:

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

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

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

4. gdeccf ...

5. 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.

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I think your method is pretty efficient. – Tunococ May 9 '13 at 22:13

Yours is I think the best way, since you retain full control over what's going on.

A variant (which I do not recommend) is that we first choose a bird of each kind, which can be done in $(20)(25)(40)(10)(5)$ ways.

Each such choice can be extended to a choice of $6$ birds in $100-5$ ways. But this overcounts, by a factor of $2$, because each "extra" bird could have been chosen in the first round or the second. So our count is $\frac{[(20)(25)(40)(10)(5)](95)}{2}$.

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Nevermind, I get it. You're just dividing by the repeats in a sequence. Thanks!!!! – user1527227 May 9 '13 at 21:10
Yes, you can think of it as $\binom{95}{1}$. Any particular choice of $1$ of each of $4$ species, and $2$ of the fifth species, could have arisen in two ways. Bald eagle Alan and bald eagle Bob can end up on the dinner plate (along with goose Charlie, duck Robert, crane Alicia, and flamingo Donna), in two ways: Alan was in the first group of $5$ shot, and Bob was shot when the intrepid hunter decided he needed a last victim, or the other way around. – André Nicolas May 9 '13 at 21:15
Thanks! Do you know if its possible to do this by thinking about it using balls and urns? – user1527227 May 9 '13 at 21:34
@user1527227: We can make a labelled funeral urn for each species, want to put $2$ in one urn, $1$ each in the others. If this urn image is helpful, OK. But to me it doesn't suggest new ways of counting. – André Nicolas May 9 '13 at 21:54
I meant composing a vector (g,d,e,c,f) and applying the multinomial theorem. In this case you want all combinations of (2,1,1,1,1) which is $\binom{5}{2,1,1,1,1}$. Since in this case we have indistinguishable balls and non-exclusivity, I thought another method would be to apply the equation $\binom{n+r-1}{r}$, but I wasn't sure how... – user1527227 May 9 '13 at 22:06

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

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Thanks for your help @Dale M. I really appreciate it. It looks like you did exactly what I did. Your method just has a fancy name ;). – user1527227 May 9 '13 at 21:06