Since the posting does not give a real answer.

192 is 11000000 in binary.

So it is simply the block with the first two bits set in the netmask.

168 is a bit more difficult. It is 10101000, a nice pattern but I don't know why this specific pattern.

I don't think this does anything to explain why 192.168/16 was chosen specifically. Three netblocks (10/8, 172.16/12, and 192.168/16) were selected from the class A, B, and C address spaces to accommodate private networks of various sizes. Class C addresses by definition have the two most significant bits set in their first octet and the third set to 0 (i.e., 192 - 223.)

192 in the first octet starts the class C space, but 10 and 172 do not have the same relationship in classes A and B.

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weinzierl 4 days ago | parent [-]

Yes you are right. I researched a bit and there are other reserved blocks next the 168 that obviously don't have a nice pattern. So the 101010 is just a coincidence.

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drewolbrich 4 days ago | parent [-]

101010 in decimal is 42.

	▲	temp0826 4 days ago \| parent [-]
		That is the answer..!

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Hikikomori 4 days ago | parent | prev [-]

192 is the first C class, 168 likely next available when rfc1918 was written.

	▲	weinzierl 4 days ago \| parent [-]
		This is the most likely thing that happened.