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MCITP MCSE : 9.  Finding the number of Hosts and subnets available for a given subnet mask: For example, let us find the number of hosts and subnets available for an IP 156.233.42.56 with a subnet mask of 7 bits. a. Class B network has the form N.N.H.H, the default subnet mask is 16 bits long. There is additional subnet mask of 7 bits long. b. 7 bits of subnet mask corresponds to (2^7-2)=128-2 = 126 subnets. c. 9 bits (16-7) of host addresses corresponds to (2^9-2)=512-2 = 510 hosts. Some times, the subnet mask is specified with the bits available in the default subnet mask. In this case the bits available in default subnet mask is 16. Therefore, total number of bits available in the subnet mask are 16+7=23. If you are given a subnet mask of 23 bits long for a class B address, it is understood that it contains the bits from the default subnet mask as well.            Hence, 126 subnets and 510 hosts are available.