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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.
  

1 Answer

+3 votes
answered by (430 points)  
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
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