Question 1180551: 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, howmany different assignments are possible
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
(1) Choose 3 of the 6 devices to be assigned IP#1; then choose 2 of the remaining 3 devices to be assigned IP#2; then choose 1 of the remaining 1 to be assigned IP#3:
C(6,3)*C(3,2)*C(1,1) = 20*3*1 = 60
OR
(2) Choose 2 of the 6 devices to be assigned IP#2; then choose 1 of the remaining 4 devices to be assigned IP#3; then choose 3 of the remaining 3 to be assigned IP#1:
C(6,2)*C(4,1)*C(3,3) = 15*4*1 = 60
OR
(3) Choose 1 of the 6 devices to be assigned IP#3; then choose 3 of the remaining 5 devices to be assigned IP#1; then choose 2 of the remaining 2 to be assigned IP#2:
C(6,1)*C(5,3)*C(2,2) = 6*10*1 = 60
As you can see from those three scenarios, the answer is the same regardless of the order in which the devices are assigned.
ANSWER: 60 different possible assignments.
----------------------------------------------------------------
You can also model the problem as the number of ways of arranging the digits 111223. For example, the arrangement 123121 would represent devices A, D, and F being assigned IP#1, devices B and E being assigned IP#2, and device C being assigned IP#3.
Then the number of possible assignments is the number of ways of arranging the digits 111223. By a well-known counting principle, that number is
|
|
|
