Path traversed during a binary search on a binary search tree is always a skewed tree.
The nodes {10,20,40,50,70,80,90} can be named as {b,b,b,b,a,a,a}.
(Here 'a' indicates after 60 and 'b' indicates before 60)
Now our work is just to find the number of permutations of the string {b,b,b,b,a,a,a}
= $\frac{7!}{4! \times 3!} = 35$
How does this work?
There is a fixed order among b's which is 10$\rightarrow$20$\rightarrow$40$\rightarrow$50
Similarly there is a fixed order among a's which is 90$\rightarrow$80$\rightarrow$70
Why this ordering is followed?
It is because we need to make our search progress towards our key=60, which is possible only by following the 2 orders: 10$\rightarrow$20$\rightarrow$40$\rightarrow$50 and 90$\rightarrow$80$\rightarrow$70.
Below I have tried some example strings.
(You can take another permutation and draw skewed tree for that permutation to have a better understanding)
Ex:
{b,b,b,b,a,a,a} => {10,20,40,50,90,80,70,(60)}
{a,a,a,b,b,b,b} => {90,80,70,10,20,40,50,(60)}
{a,b,a,b,b,b,a} => {90,10,80,20,40,50,70,(60)}
Hope this helps! :)
Let me know if there is any mistake here!