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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?

Mar 3, 2020

#1
+1

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   Mar 3, 2020
#2
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Sorry that is not the correct answer

Guest Mar 3, 2020