how many subnets are there in a class C address 192.168.3.0/24?

Answer 1

You can split the network many ways. Forget about the a.b.c portion of the network address and only deal with the last byte.
You have 256 binary values to deal with. For this network, .0 is the network address, and .255 is the broadcast address. One of the addresses needs to be the default gateway.

For every subnet 3 addresses are used up by the network address, the default gateway and the broadcast address. this requires 2 bits of overhead (1.5 bits) in the network address for subnets. Subnets of 2 bits containing 1 usable IP address are generally not worth creating for any reason.

In theory, you could break the 8 bits into 64 - 2 bit groups.
To address 64 different subnets, you need 6 bits, for the network address. Not all of those groups would be usable because the first and last of those addresses are already being used by the parent subnet for the network address, default gateway and broadcast address, so 62 would be usable.
(.0 = network, .254 = gateway, .255 = broadcast.)

That would make 62 usable subnets plus the class C subnet for a theoretical maximum of 63.

In the actual world you need to have a subnet set aside that has enough usable IP addresses for each of the routers that will control a subnet. Lets break the last byte into to groups of bits r and u where r stands for routers and u stands for useable IP address.

The subnet as it stands is uuuuuuuu (except for the network, broadcast and default gateway addresses) .

It could be broken up into:
ruuuuuuu = 2 subnets of 128 IP addresses.
rruuuuuu = 4 subnets of 64 IP addresses.
rrruuuuu = 8 subnets of 32 IP addresses.
rrrruuuu = 16 subnets of 16 IP addresses.
rrrrruuu = 32 subnets of 8 IP addresses.
rrrrrruu = 64 subnets of 4 IP addresses.

Since one of the subnets defined needs to be used for the routers, at least one subnet needs to be created that can hold enough values for the r's.

An example of breaking down the Class C into smaller subnets:

The class C subnet has a default gateway router with an IP address lets say of 254. We could break the class C into smaller groups of lets say 16 IP addresses.
The network addresses of these subnets would be multiples of 16, x.x.x.0 through x.x.x.224. with a 28 bit subnet mask. The subnet containing the routers to those subnets would be an exception. Its address would be x.x.x.224 with a 27 bit subnet mask.
Here is how the subnets would look.

x.x.x.0/24 = Class 3 subnet it has a network address of x.x.x.0 with a broadcast address of x.x.x.255 and a default gateway of x.x.x.224

(outside world) -> 192.168.3.0/24 -> (gateway external IP address)

gateway internal address 192.168.3.254 -> 192.168.3.224/27 subnet

The routers on the 192.168.3.224 subnet could have any 15 of the IP addresses in the range of 192.168.3.225 through 192.168.3.253. The other unused IP addresses in that range could be used by computers or printers. If the first 15 IP addresses were used, this is what the routing tables would look like.
Destination via Gateway
192.168.3.0/28 <-> 192.168.3.225
192.168.3.16/28 <-> 192.168.3.226
192.168.3.32/28 <-> 192.168.3.227
192.168.3.48/28 <-> 192.168.3.228
192.168.3.64/28 <-> 192.168.3.229
192.168.3.80/28 <-> 192.168.3.230
192.168.3.96/28 <-> 192.168.3.231
192.168.3.112/28 <-> 192.168.3.232
192.168.3.128/28 <-> 192.168.3.233
192.168.3.144/28 <-> 192.168.3.234
192.168.3.160/28 <-> 192.168.3.235
192.168.3.176/28 <-> 192.168.3.236
192.168.3.192/28 <-> 192.168.3.237
192.168.3.208/28 <-> 192.168.3.238

192.168.3.239 -> 192.168.3.253 would be usable IP addresses on the 192.168.3.224/27 network.
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Explanation of one of those subnets.
The 192.168.3.208/28 subnet would have a network address of 192.168.2.208, a broadcast address of 192.168.3.223 and could have a default gateway of 192.168.3.209. A PC with an address of 192.168.3.210 on that network connecting to the internet would send its traffic through its default gateway of 192.168.3.209. The router with this IP address on the 192.168.3.208/28 network would also have the IP address of 192.168.3.238 on the 192.168.3.224/27 network. That router would send the traffic to the internet through its default gateway of 192.168.3.254. It is not specified what the external IP address of the 192.168.3.0 subnet is. When a packet came in from the internet to the address of 193.168.3.210, first the IP address would be masked against 255.255.255.0 (the netmask of the 192.168.3.0 network). Since the address matched this class C network, the traffic would come in through the default gateway onto the 192.168.3.224 network. There, the netmask of 255.255.255.240 (28 bit) is applied to the address and the resulting number is 192.168.3.208. The routing table states that the gateway to the 192.168.3.208 network is the 192.168.3.238 router. The traffic is sent to that router. That router has an IP address on the 192.168.3.208/28 network of 192.168.3.209. It sends the traffic back to the PC with an IP address of 192.168.3.210 on that network.
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That would make 14 /28 bit subnets with 13 usable IP addresses each and 1 /27 bit subnet with 15 remaining usable IP addresses for a total of 15 subnets broken out of the class C subnet. If you include the class C subnet, a total of 16 subnets would have been created.

Since the available IP addresses in the x.x.x.224/27 subnet could be used to split the /28 bit subnets into smaller subnets, an additional 15 subnets could be created for a total of 31 subnets.

For example the definition of
192.168.3.0/28 <-> 192.168.3.225
could be split into
192.168.3.0/29 <-> 192.168.3.225
192.168.3.8/29 <-> 192.168.3.239

That would increase the subnet count, but reduce the number of useable IP addresses by 4.

Answer 2

Or the short version. Determine the network which is 24 bits. Determine the Prefix which is also the 24 bits because you add the Network bits (24) plus the subnet bits (0) to get your Prefix. Determine your subnet bits by Subtracting the Network bits (24) from the Prefix bits (24). That yields 0 subnet bits. The total subnets is determined by calculating 2 to the power of the subnet bits. In this case we have 0 subnet bits. Any number to the power of 0 is 1. So 2 to the power of 0 is 1. You have 1 subnet for /24. Then you will recognize the binary pattern from left to right. 1 subnet for /24, 2 subnets for /25, 4 subnets for /26, 8 subnets, 16 subnets, 32, subnets, 64 subnets for /30 (point to point), and then you run out of hosts.

Answer 3

By definition, a class C network uses the first three sets of numbers to identify the network, and the last set to identify the node (individual computers). So since there is only one set of numbers available for nodes, it can not be split into a subnet. You do that by varying the third set of numbers.

So a Class C can only have one subnet.