Class C sub netting

IP address 198.100.98.0
Decimal mask 255.255.255.224
Convert 224 into binary -> 11100000(2)
Maximum no. of combinations = 2n (Where n is number of masked bits)
Note: Masked bit = 1, UN masked bit= 0.
Maximum number of subnets -> 2n-2
Maximum number of hosts/subnet -> 2m-2 (Where m is number of unmasked bits)
Now 224(10) = 11100000(2)
Maximum no. of combinations = 2n = 23 = 8
Maximum no. of subnets =2n-2 = 23-2 = 6
Maximum no. of hosts/subnet=2m -2 =25 -2 = 30

Available combinations

0 -> 000 (This is not used for assigning IP address)
1 -> 001
2 -> 010
3 -> 011
4 -> 100
5 -> 101
6 -> 110
7 -> 111 (This is not used for assigning IP address)
Octant contains 8 bits, presently we have 3 bits (Remaining 5 bits are zeros)
00100000 = 32
01000000 = 64
01100000 = 96
10000000 = 128
10100000 = 160
11000000 = 192
Therefore modified subnets are
255.255.255.32
255.255.255.64
255.255.255.96
255.255.255.128
255.255.255.160
255.255.255.192