[Math] $4$ integers are randomly selected from the numbers from $1$ to $10$. The chance that there are at least two successive numbers among those $4$ is

Question

$4$ integers are randomly selected from the numbers from $1$ to $10$. The chance that there are atleast two successive numbers among those $4$ selected is
$(A)\frac{5}{6}\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{2}{3}\hspace{1cm}(D)\frac{1}{2}\hspace{1cm}$

I calculated answer as $\frac{24}{\binom{10}{4}}$ but this wrong. Please help me find the right answer.

Answer 1

Take out $4$ numbers, 6 numbers (N) are left with $7$ possible gaps including ends

$_ N _ N _ N _ N _ N _ N _$

We can replace the $4$ numbers in forbidden way in any of 7 gaps (including ends) in $\dbinom{7}{4}$ ways

thus indicated $Pr = 1 - \dfrac{\dbinom{7}{4}}{\dbinom{10}{4}}$