Each subnet mask has 32 bits which are sperate in network and host bits. In this case we have 28 network bits which means 32 - 28 = 6 Host bits remain. 2^6 = 64.. BUT every subnet has 1 network adress (the first IP address in that subnet) and 1 broadcast address (the last one in that subnet) so 64 -2 = 62 hosts in each subnet
Correct. IP for this subnet is 192.168.2.128 ~ .191.
However, you need to remove the subnet address (.128) and broadcast address (.191)
...and you are left with 62 unique host addresses.
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