# Sticks and stones problem

Ten squares in a row are labelled 1, 2, 3, . . . 10, in order. A counter starts at square 1. At every step, the counter can move ahead 1, 2, or 3 squares. However, beginning with the second step, the counter cannot move the same number of squares as it did in the previous step. For example, in the first step, the counter can move from square 1 to square 3. Then in the next step, the counter can move to square 4 or square 6, but not square 5. Find the number of possible sequences of steps that take the counter from square 1 to square 10. (In the last step, the counter must land exactly on square 10.)

Hi, I am thinking of using sticks and stones for this problem, but cannot think of a way to use it. Can anyone help me?

Thanks!

• Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem; this will prevent people from telling you things you already know, and help them give their answers at the right level. – JKnecht Feb 17 '16 at 5:30
• I think a computer will work better than sticks and stones. – Gerry Myerson Feb 17 '16 at 6:17
• I mean the method, sticks and stones :) – Brad Feb 17 '16 at 6:31
• I think a computer will work better than the method, sticks and stones. – Gerry Myerson Feb 17 '16 at 12:12

I get $21$.

$2$ with no $3$'s:

$1,2,1,2,1,2$
$2,1,2,1,2,1$

$12$ with one $3$:

$3,1,2,1,2$
$3,2,1,2,1$
$1,3,2,1,2$
$2,3,1,2,1$
$1,2,3,1,2$
$1,2,3,2,1$
$2,1,3,1,2$
$2,1,3,2,1$
$1,2,1,3,2$
$2,1,2,3,1$
$1,2,1,2,3$
$2,1,2,1,3$

$7$ with two $3$'s:

$3,1,2,3$
$3,2,1,3$
$1,3,2,3$
$3,2,3,1$
$1,3,1,3,1$
$2,3,1,3$
$3,1,3,2$

$2+12+7=21.$