On a Class B network, using a subnet mask with an octet value of 248 (i.e., 255.255.248.0 → borrowing 5 bits in the third octet), how many subnets are available under the traditional classful rule set?

Introduction / Context:
Before CIDR and modern practices, classful subnetting calculated the number of subnets based on bits borrowed from the host portion and often applied the “no all-zeros, no all-ones” rule to subnet IDs. This question uses that legacy approach.

Given Data / Assumptions:

• Base class: Class B (default /16).
• Subnet mask: 255.255.248.0 → /21 (i.e., 16 + 5 bits borrowed).
• Legacy rule: subtract 2 subnets (all-zeros and all-ones subnets not used).
• We are counting subnets, not hosts per subnet.

Concept / Approach:
Borrowed bits = 5 (since 248 = 11111000 in the third octet). Under the traditional rule, number of subnets = 2^(borrowed bits) − 2 = 2^5 − 2 = 30. (Modern networks typically allow using all subnets, giving 32, but classical exam questions often expect the legacy 30.)

Step-by-Step Solution:
Determine borrowed bits: from /16 to /21 → 5 bits.Compute classic count: 2^5 − 2 = 30 subnets.Note: Hosts per /21 subnet = 2^(32 − 21) − 2 = 2046 usable (not asked, but useful context).

Verification / Alternative check:
Binary increment of the third octet in steps of 8 (since 248 leaves 3 host bits) shows distinct subnet IDs: .0, .8, .16, …, up to .248, which corresponds to 32 possible patterns; subtracting the two special patterns yields 30 subnets.

Why Other Options Are Wrong:
2 or 6: Far too small; borrowing 5 bits yields many more subnets.

62 or 126: These match subtract-2 results for borrowing 6 or 7 bits, not the 5-bit case.

Common Pitfalls:
Mixing modern “use all subnets” practice (32) with the legacy subtract-2 convention (30), and confusing host-bit math with subnet-bit math.

Final Answer:
30