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

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

Explanation

Solution

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

Complete step-by-step answer:
Let us consider the right hand side:
RHS=n−1Pr+r. (n−1)Pr−1RHS = {}^{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=n!(n−r)!{}^nPr = \dfrac{{n!}}{{\left( {n - r} \right)!}}
Applying this formula we get:-
RHS=(n−1)!(n−1−r)!+r.(n−1)!(n−1−(r−1))!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=(n−1)!(n−1−r)!+r.(n−1)!(n−1−r+1)! RHS=(n−1)!(n−1−r)!+r.(n−1)!(n−r)!  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:
(n−r)!=(n−r)(n−1−r)!\left( {n - r} \right)! = \left( {n - r} \right)\left( {n - 1 - r} \right)!
Hence substituting the value we get:-
RHS=(n−1)!(n−1−r)!+r.(n−1)!(n−r)(n−r−1)!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 (n−1)!(n−1−r)!\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - r} \right)!}} common we get:-
RHS=(n−1)!(n−1−r)![1+r.1n−r]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=(n−1)!(n−1−r)![1(n−r)+r(1)n−r] RHS=(n−1)!(n−1−r)![n−r+rn−r] RHS=(n−1)!(n−1−r)![nn−r]  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(n−1)! (n−r)!=(n−r)(n−1−r)!  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=n!(n−r)!RHS = \dfrac{{n!}}{{\left( {n - r} \right)!}}
Also, we know that:
nPr=n!(n−r)!{}^nPr = \dfrac{{n!}}{{\left( {n - r} \right)!}}
Therefore,

RHS=nPr  = LHS  RHS = {}^nPr \\\ {\text{ = }}LHS \\\

Therefore,
RHS = LHSRHS{\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.