You have 192.168.10.0/24 and 10 departments:
Firts, i will suppose that you don' t have any requirement about your address space, because i will give you the basic answer.
What you need to do its subnetting looking 10 different networks>
You need to remember that there is a rule for that: (2^n) where n is the number of bits
If you take 3 bits: 8 networks, NOT
If you take 4 bits: 16 networks, YES
192.168.10.0000 0000 the last octet it is in bit notation
192.168.10.0000 0000
192.168.10.0001 0000
192.168.10.0010 0000
192.168.10.0011 0000
192.168.10.0100 0000
192.168.10.0101 0000
192.168.10.0110 0000
192.168.10.0111 0000
192.168.10.1000 0000
192.168.10.1001 0000
192.168.10.1010 0000
192.168.10.1011 0000
192.168.10.1100 0000
192.168.10.1101 0000
192.168.10.1110 0000
192.168.10.1111 0000
The part of the octet that i has write in bold form, that bits now will be part of network, and ther your mask will move to 28.
On that way, you have your 10 required networks and 6 networks more.
Then your network will have 2^n -2 host, cause the first address is your networks address and the last is your broadcast address. Every network will have then 14 hosts.