Six devices (A,B,C,D,E, F) are each assigned one of three IP addresses...how many different assignments are possible?

Question

Six devices (A,B,C,D,E, F) are each assigned one of three IP addresses (IP#1, IP#2, IP#3). If three
devices are assigned IP#1, two devices are assigned IP#2, and one device IP#3, how many different
assignments are possible?

This is question 4 from lecture 3A.

Answer 1

The first step is to find out the number of ways to assign IP#1 to 3 devices out of all the devices. Since at ths initial stage there are 6 devices, then there are C(6,3) or 20 ways.

remember that C(n,r) = n!/r!(n-r)!

The next step is to find out how my ways to assign IP#2 to 2 devices out of the remain 3 devices. There are C(3,2) or 3 ways to do that.

Then lastly finding the number of way to assign IP#3 to the last remaining device, there is only 1 way to do that (C(1,1)).

Now by the Basic Counting Rule the total number of assignments possible is C(6,3)*C(3,2)*C(1,1) or 20*3*1 which is 60 total assignments