Let $f(n)$ be the number of the binary search trees of $(n-1)$ height (i.e. maximum height) using numbers $(x+1,x+2,x+3,...,x+n)$ where $x \in \mathrm{R}$.
For $n=1$, the answer is $1$ since there is only one node of height $0$. $\therefore f(1)=1$
For $n=2$, the answer is $2$ since there are two cases of height $1$ starting $2$ as root and $1$ as root . $\therefore f(2)=2$
For $n=3$, the answer is $4$.
Let's observe how $f(n)$ is growing from $f(n-1)$. We can add the $n^{\mathrm{th}}$ node taking as the root to the trees of $(n-1)$ nodes $(1,2,3,...,n-1)$. For this case of $(n-1)$ nodes, there are $f(n-1)$ trees. Still there are more cases to be taken. Take $1$ as the root and the rest $(n-1)$ nodes are $(2,3,...,n)$ for which there are also $f(n-1)$ trees. [ For example, we can add $3$ as the root to trees of two nodes $(1,2)$. Besides taking $1$ as the root, there can be trees using $(2,3)$. That's why $f(3)=f(2)+f(2)=2+2=4$ ]
So $f(n)=\{\text{trees using }(1,2,3,...,n-1)\text{ nodes having }n^{\mathrm{th}}\text{ node as the root}\}+\{\text{trees using }(2,3,4...,n)\text{ nodes having 1 as the root}\}$
$\begin{align}\therefore f(n)&=f(n-1)+f(n-1)\\ &=2f(n-1)\\&=2^2f(n-2)\\&=2^3f(n-3)\\&=\cdots\\&=2^{n-1}f(1)\\&=2^{n-1}; ~[\because f(1)=1] \end{align}$
$\therefore f(7)=2^{7-1}=2^6=64$.
So the correct answer is $64$.