The address of class B host is to be split into subnets with 6‐bit subnet…

Solution:

• Start with Class B: default mask is /16, so there are 16 host bits.

• Borrow 6 bits for the subnet number. Number of possible subnets = 2^6 = 64.

• Remaining host bits = 16 - 6 = 10. Total addresses per subnet = 2^10 = 1024.

• Usable hosts per subnet = total addresses minus network and broadcast = 1024 - 2 = 1022.

• If following the historical restriction that disallowed subnet-zero and the all-ones subnet, subtract 2 from the subnet count: 64 - 2 = 62 usable subnets. If subnet-zero and all-ones are allowed (modern practice), you have all 64 subnets available.

Final answer: 62 subnets and 1022 hosts (using the older convention of excluding subnet-zero and all-ones). If subnet-zero and all-ones are allowed, the subnet count would be 64 and usable hosts per subnet remain 1022.