Good Parking Spaces

Math: Permutations: 15 parking spaces, 4 cars, no 2 cars can be next to each other...?

There are 15 parking spaces. 4 different cars are parking in them. In how many ways can the four cars park IF NO TWO CARS PARK NEXT TO EACH OTHER?

Public Comments

1. If you label the car park spaces A to 0 you have ACEG ACEH ACEI ACEJ ACEK ACEL ACEM ACEN ACEO then ACFH ACFI ACFJ ACFK ACFL ACFM ACFN then ACGI ACGJ ACGK ACGL ACGM ACGN then ACHJ ACHK ACHL ACHM ACHN then ACIK ACIL ACIM ACIN then ACJL ACJM ACJN then ACKM ACKN then ACLN so a total of 37. Then repeat the process for ADFH etc, AEFI etc then move on to the Bs. Very long winded way of doing it so I'm sure someone will come up with a faster way by using some kind of equation. lol.
2. Parking of 4 cars is like as #A#B#C#D# A,B,C,D are 4 different cars # are available spaces 4 different cars use 4 spaces and there are 11 available spaces remain. Cannot use #1+#2+#3+#4+#5 = 11 because #1 and #5 can be 0 then must adapt to be #1+#2+#3+#4+#5 = 13 ; where #1,#2,#3,#4,#5 > 0 possible ways of available spaces is (13-1)C(5-1) = 12C4 and 4 different cars can be permutations in 4! different ways so the answer is 4!*12C4 ANSWER