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If ${}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{r}}$ then find the value of r.

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Last updated date: 19th Jul 2024
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Answer
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Hint: Use the fact that ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ and $0\le r\le n$. Calculate the values of ${}^{9}{{P}_{5}}$ and ${}^{9}{{P}_{4}}$ ,hence determine the value of LHS. Now start by substituting r = 0,1,2,3, … till LHS = RHS.

Complete step-by-step answer:
$\begin{align}
  & {}^{9}{{P}_{5}}=\dfrac{9!}{\left( 9-5 \right)!}=\dfrac{9!}{4!}=\dfrac{9\times 8\times 7\times 6\times 5\times 4!}{4!}=9\times 8\times 7\times 6\times 5=15120 \\
 & {}^{9}{{P}_{4}}=\dfrac{9!}{\left( 9-4 \right)!}=\dfrac{9!}{5!}=\dfrac{9\times 8\times 7\times 6\times 5!}{5!}=9\times 8\times 7\times 6=3024 \\
\end{align}$
Hence LHS $=15120+5\times 3024=30240$.
When r = 0, RHS = $\dfrac{10!}{\left( 10-0 \right)!}=1$
When r = 1, RHS = 10
When r = 2 RHS = $10\times 9=90$
When r = 3 RHS = $90\times 8=720$
When r = 4 RHS = $720\times 7=5040$
When r = 5 RHS = $5040\times 6=30240$
Hence, we have when r = 5 LHS = RHS.
Hence r = 5.

Note: Proof by argument:
Number of Permutations of 10 Letters taken 5 at a time = Number of permutations if a particular letter (say A) always occupies the first position + Number of permutations if that particular letter never occupies the first position.
LHS = \[{}^{10}{{P}_{5}}\].
The number of permutations if a particular letter (say A) always occupies the first position = Number of permutations of remaining 9 letter taken 4 at a time = \[{}^{9}{{P}_{4}}\]
The number of permutations if a particular letter never occupies the first position = \[9\times {}^{9}{{P}_{4}}\] Because the first place has 9 choices [NOT TAKING THE PARTICULAR LETTER] and the remaining letters can be arranged in \[{}^{9}{{P}_{4}}\] ways.
Now we know that $\left( n-r+1 \right){}^{n}{{P}_{r-1}}={}^{n}{{P}_{r}}$
Hence, we have
$\begin{align}
  & \left( 9-5+1 \right){}^{9}{{P}_{4}}={}^{9}{{P}_{5}} \\
 & \Rightarrow 5{}^{9}{{P}_{4}}={}^{9}{{P}_{5}} \\
\end{align}$
Hence \[9{}^{9}{{P}_{4}}=5{}^{9}{{P}_{4}}+4{}^{9}{{P}_{4}}={}^{9}{{P}_{5}}+4{}^{9}{{P}_{4}}\]
Hence, we have
\[\begin{align}
  & {}^{10}{{P}_{5}}={}^{9}{{P}_{4}}+{}^{9}{{P}_{5}}+4{}^{9}{{P}_{4}} \\
 & \Rightarrow {}^{10}{{P}_{5}}={}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}} \\
\end{align}\]