If P (5, r) = P (6, r − 1),

$P(5, r)=P(6, r-1)$

or ${ }^{5} P_{r}={ }^{6} P_{r-1}$

$\frac{5 !}{(5-r) !}=\frac{6 !}{(6-r+1) !}$

$\Rightarrow \frac{(6-r+1) !}{(5-r) !}=\frac{6 !}{5 !}$

$\Rightarrow \frac{(7-r) !}{(5-r) !}=\frac{6(5 !)}{5 !}$

$\Rightarrow \frac{(7-r)(6-r)(5-r) !}{(5-r) !}=6$

$\Rightarrow(7-r)(6-r)=6$

$\Rightarrow(7-r)(6-r)=3 \times 2$

On comparing the above two equations, we get:

$7-r=3$

$\Rightarrow r=4$