Sometimes, jumping 3 at a time can change your face!

Since I'm not so good at combinatorics, I chose to do it in a no-brainer method, i.e., by simple counting.

Hence, total ways to do it is $=1+8+15+4=28$

We can form an equation as $a + 3b = 10$ . So, we obtain $4$ solutions ;

$b = 0 , a = 10$

$b = 1 , a = 7$

$b = 2 , a = 4$

$b = 3 , a = 1$

Here the values of $b$ indicate the number of steps in which he climbed up $3$ stairs and $a$ indicates the number of steps in which he climbed up $1$ stair.

So for case $I$ we have $10$ steps of $1$ stair hence permutations $= \dfrac{10!}{10!\times0!} = 1$

For case $II$ we have $7$ steps of $1$ stair and $1$ step of $3$ stairs hence permutations $= \dfrac{8!}{7!\times1!} = 8$

Similarly for cases $III$ and $IV$ we have permutations $= \dfrac{6!}{4!\times2!} = 15$ and $\dfrac{4!}{3!\times1!} = 4$

Adding all the cases : $1 + 8 + 15 + 4 = \boxed{28}$