Prove that \[{}^nPr = {}^{n - 1}Pr + r.{\text{ }}{}^{\left( {n - 1} \right)}Pr - 1\].

Hint: Here we will first consider the RHS and then prove it equal to LHS by using the following formula and simple arithmetic operations:
\[{}^nPr = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]

Complete step-by-step answer:
Let us consider the right hand side:
\[RHS = {}^{n - 1}Pr + r.{\text{ }}{}^{\left( {n - 1} \right)}Pr - 1\]
Now applying the following formula separately for each term we get:-
The formula is:
\[{}^nPr = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Applying this formula we get:-
\[RHS = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}} + r.\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - \left( {r - 1} \right)} \right)!}}\]
Now solving it further we get:-
\[
  RHS = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}} + r.\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r + 1} \right)!}} \\
  RHS = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}} + r.\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - r} \right)!}} \\
 \]
As we know that:
\[\left( {n - r} \right)! = \left( {n - r} \right)\left( {n - 1 - r} \right)!\]
Hence substituting the value we get:-
\[RHS = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}} + r.\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - r} \right)\left( {n - r - 1} \right)!}}\]
Now taking \[\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}}\] common we get:-
\[RHS = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}}\left[ {1 + r.\dfrac{1}{{n - r}}} \right]\]
Now taking the LCM and solving it further we get:-
\[
  RHS = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}}\left[ {\dfrac{{1\left( {n - r} \right) + r\left( 1 \right)}}{{n - r}}} \right] \\
  RHS = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}}\left[ {\dfrac{{n - r + r}}{{n - r}}} \right] \\
  RHS = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}}\left[ {\dfrac{n}{{n - r}}} \right] \\
 \]
Now we know that:-
\[
  n! = n\left( {n - 1} \right)! \\
  \left( {n - r} \right)! = \left( {n - r} \right)\left( {n - 1 - r} \right)! \\
 \]
Hence substituting the values we get:-
\[RHS = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Also, we know that:
\[{}^nPr = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Therefore,
\[
  RHS = {}^nPr \\
  {\text{ = }}LHS \\
 \]
Therefore,
\[RHS{\text{ = }}LHS\]
Hence proved.

Note: Students should proceed from Right hand side in such questions as it is much easier to compress two or more terms into a single term rather than expanding a single term into two or more terms.
The formula of permutation should be used with correct values of n and r.