Question

If \[{}^{9}{{P}_{5}}+5\cdot {}^{9}{{P}_{4}}={}^{10}{{P}_{r}}\] , find value of r.

Answer 1

Hint: We will use the formula of permutation which is given as \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\] and putting this formula, we will expand the given equation \[{}^{9}{{P}_{5}}+5\cdot {}^{9}{{P}_{4}}={}^{10}{{P}_{r}}\] . After expanding, we will make the LHS side equal to RHS and then by comparing we can get the answer.

Complete step-by-step answer:
Here, we are given equation \[{}^{9}{{P}_{5}}+5\cdot {}^{9}{{P}_{4}}={}^{10}{{P}_{r}}\] and we have to find value of r.
So, we will use formula of permutation which is given as \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\] . We will get as
\[\Rightarrow \dfrac{9!}{\left( 9-5 \right)!}+5\cdot \dfrac{9!}{\left( 9-4 \right)!}=\dfrac{10!}{\left( 10-r \right)!}\]
On solving, we get
\[\Rightarrow \dfrac{9!}{4!}+5\cdot \dfrac{9!}{5!}=\dfrac{10!}{\left( 10-r \right)!}\]
Now, we can write \[\dfrac{5}{5!}=\dfrac{5}{5\times 4!}=\dfrac{1}{4!}\] so, we will get
\[\Rightarrow \dfrac{9!}{4!}+\dfrac{9!}{4!}=\dfrac{10!}{\left( 10-r \right)!}\]
On further simplification, we can get equation as
\[\Rightarrow \dfrac{2\cdot 9!}{4!}=\dfrac{10!}{\left( 10-r \right)!}\]
Now, we will multiply 5 on the LHS side to make it the same as that of RHS.
\[\Rightarrow \dfrac{2\cdot 5\cdot 9!}{5\cdot 4!}=\dfrac{10!}{\left( 10-r \right)!}\]
On solving we get
\[\Rightarrow \dfrac{10!}{5!}=\dfrac{10!}{\left( 10-r \right)!}\]
So, on comparing the equation we can write denominator as
$\Rightarrow 5=10-r$
$\Rightarrow r=10-5=5$
Thus, the value of r is 5.

Note: Do not make mistakes in using the formula of permutation because it is slightly changed than the combination formula i.e. $\dfrac{n!}{\left( n-r \right)!\cdot r!}$ . While using this formula, the whole can get changed or sometimes it results in decimal form which is not true. But students should know that the value of r will be an integer. So, do not make this mistake.