Five Texans in their five distinct large pickup trucks want to park in an empty row of fifteen undersized California parking spaces. Because each truck just barely fits in one parking space, the Texans want to leave at least one empty space between their trucks, and at least one empty space at each end of the row. In how many different ways can they park in the row?
please help!!!! any help is greatly appreciated!!! :)
ps stay safe guys!
I am not sure but this is what comes to mind. Draw it as you go, so you can picture what I mean.
The row started with an E (for empty.) Then within the row there are 5 pairs of (Truck,Empty)
So that accounts for 11 spaces. there are 4 spaces left to fill
Some, or all could be on either side of the allotted pairs.
So there are 6 places they can go. (they can all be in the same place)
So there are 5 bars. You can draw these through the middle of the circled TE pairs.
The 4 spaces are represented by stars (stars and bars method)
So I think I need to choose 4 (or 5 which is the same ) from 9
9C5=126
oh you were so close! you got everything right except that you have to multiply 126 by 120 because there are 5!=120 ways to permute the trucks. so the answer according to the website is 126*120=15120
thanks for the help though! i really appreciate it! :)