I also found two incorrect answers about subnetting one of wich has been corrected. The other one is this:
Question A24: you plan to set up 4 subnets with up to 20 host on each. You'll use 192.168.0.0/24 private addr. Which subnet mask should you assign?
a. 255.255.255.192
b. 255.255.255.224

What is your suggestion? And why?

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I would to B, because 2^3-2=6 subnets which meets the minimum requirement of 4.

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I would choose a. because 2^2=4.
(When calculating subnets you don't subtract 2 only when claculating hosts.)
You need 2 bits from the 4. octet for creating 4 subnets.

The subnets:
192.168.0.0/26 (00)
192.168.0.64/26 (01)
192.168.0.128/26 (10)
192.168.0.192/26 (11)

Each subnet gives 62 hosts
The Transcender's answer was b. by the way
which is wrong IMHO. They probably calculated the number of hosts first and then the subnets.

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Microsoft doesn't want you to consider the all 0s and all 1s subnets as valid, even though they are valid in the real networking world. So therefore you will need to subtract two from the number of subnets in order to get it right on the exam.

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I have done some exams and never seen what you are talking about. If you can create 4 subnets in reality from 2 bits then it must be valid in the exams too.
When you calculate subnets you don't subtract two only when calculate hosts.
All 0's in the host portion is the (sub)network address, 255 is the (sub)network loopback address, and 256=all 1's is the next (sub)network address. That's why you have to subtract 2.

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Almost forgot: there are no all 0's and all 1's subnets.
Or if you consider 192.168.1.0 as all zero than yes there are, but no need subtracting when creating subnets.

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I suggest you review Freak's subnetting guide for more information. I'm not just making this stuff up for the sake of argument.

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you actually did give any reason why I am wrong.

I give you another hint: you need only two subnets. According to your logic 2^1=2-2=0.

According to my logic 2^1=2 i.e you can create 2 subnets with two bits, I mean I can.
Can you?

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Sure you can create two subnets using 2 bits, but you cannot create 2 subnets using 1 bit. For example, 192.168.1.0 with a mask of 255.255.255.128. Available subnets are 192.168.1.0 and 192.168.1.128. The first one contains all 0s and is the network ID and the second one contains all 1s and is the broadcast ID.

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RFC 950 originally defined the all 0s and all 1s subents as invalid, because they could be confused between classful and classless routing. However, since routers can now support both types of routing there is no need to eliminate 2 of those subnets.

I'm pretty sure that M$ still wants you to subtract those 2 subnets just in case you administer an environment where CIDR is not supported.

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