Suppose that there are 15 antennas in a store, of which 3 are defective. Assume all the defectives and all the functional antennas are indistinguishable. If we lay all the antennas down in a row, how many linear orderings are there in which no two defectives are consecutive?
+128472 1 2 3
...
13 14 15 = 13 orderings where all 3 are together
1 2
....
14 15
There are 14 positions where any two are together and for any of these the other defective could occupy any one of 12 positions
So......the number of orderings where 2 are together = 14 * 12 = 168
The total number of identifiable orderings = 15! / [ 3! * 12! ] = 455
So....the number of orderings wher no two defectives are together =
Total identifiable orderings - orderings where all three are together - orderings where two are together =
455 -13 - 168 =
274
