Hello student,
Please find the answer to your question below
There are6possibilities if the last three stops are at8,10, and12. Now leave the last two stops at10and12and move the second stop forward to7: there are just5possibilities for the first stop. If you move the second stop forward to6, there are4, and so on, with just one when you move it all the way to3; these cases give you a total of6+5+…+1=6⋅72=21possibilities. Now move the third stop forward to9; the same analysis will show that you have5+…+1=5⋅62=15possibilites. As you keep moving the third stop forward, you get4+3+2+1=10,3+2+1=6,2+1=3, and1possibility, for a total at this point of21+15+10+6+3+1=56possibilities. Now repeat with the fourth stop moved forward to11. You’ll quickly find that most of the counting is a repetition of what you’ve already done, and you end up with15+10+6+3+1=35possibilities. Similarly, with the last stop at10you end up with10+6+3+1=20possibilities. In the end you find yourself adding up 56+35+20+10+4+1to get126.