It totally depends on the subnet mask. If you're talking about a 255.255.0.0 (/16 CIDR), you get (255*255)-1 hosts - 65024. If you're using 255.255.255.0 (/24 CIDR) masks and individual subnets, you get 254*255=
64,770.
For a simple class B network, which has a default mask of 255.255.0.0, would have 16 bits for hosts, so 2^16=65536. You minus the invalid hosts, the network and broadcast address, and your left with 65534 valid hosts.
I made a mistake in my math - I forgot that in a class B, all the ".0" IP's are valid as addresses (except the first one). 2^16th-2 is indeed the correct answer.
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