Counting pairs of animals 
Old MacDonald has $5$ chickens, $4$ donkeys, and $7$ emus. How many ways can he pair up the animals so that every pair consists of animals of different species? (The order of the animals within each pair does not matter, and the order among the pairs does not matter. Assume that all animals are distinguishable.)

The answer I have currently is $83$ since there are $5*4=20$ possible pairs for (Chicken, Donkey), $5*7=35$ possible pairs for (Chicken, Emu), $4*7=28$ possible pairs for (Donkey, Emu). Then I added them up: $20+35+28=83$. Is this correct?
 A: Correct - you get the same thing by counting all possible pairs and subtracting the ones with two of the same type ..

A: Yup, your solution is correct.

Making this a community wiki post since it's not like I have anything to add. If anyone else does, they're welcome to, though!
A: The answers given so far are wrong: they count the number of possible pairs but not the number of ways to pair all the animals up at once. (We can think of these as matchings.)
Notice that the 7 emus must be paired with 4 chickens and 3 donkeys, leaving 1 chicken and 1 donkey to form a pair. Any other option (e.g. pairing the emus with 5 chickens and 2 donkeys) leaves unequal numbers of chickens and donkeys, who then can't be paired up.
Each valid matching is specified in exactly one way by a sequence of choices as follows:

*

*Divide the emus into a set of 3 and a set of 4. There are $\binom73=\binom74=35$ ways to do this.

*Pair the emus from the set of 3 with donkeys one at a time: $4\times3\times2=24$ ways.

*Pair the emus from the set of 4 with chickens: $5\times4\times3\times2=120$ ways.

*Pair the remaining donkey and chicken together (1 way).

So the total number of ways to pair the animals up is $35\times24\times120=100800$.
