We are to form 4-digit numbers with either 5 or 6 or 7 or 8 or 9 at the thousand's place. 

4-digit number :     $\begin{array}{l}X\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}X\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}X\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}X\\ \downarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\downarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\downarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\downarrow \\ 1 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\hspace{0.25em} \hspace{0.25em}4\end{array}$

No. of ways of filling 1st place = 5 After filling 1st -place, we have 8 digits to be used. 

No. of ways of filling 2 nd , 3 rd, 4 th places = ${ }^8{P}_3$

$\therefore$Total no. of numbers formed $=5{\times }^8{P}_3$