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

If 9P5+5⋅9P4=10Pr{}^{9}{{P}_{5}}+5\cdot {}^{9}{{P}_{4}}={}^{10}{{P}_{r}} , find value of r.

Explanation

Solution

Hint: We will use the formula of permutation which is given as nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} and putting this formula, we will expand the given equation 9P5+5⋅9P4=10Pr{}^{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 9P5+5⋅9P4=10Pr{}^{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 nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} . We will get as
⇒9!(9−5)!+5⋅9!(9−4)!=10!(10−r)!\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
⇒9!4!+5⋅9!5!=10!(10−r)!\Rightarrow \dfrac{9!}{4!}+5\cdot \dfrac{9!}{5!}=\dfrac{10!}{\left( 10-r \right)!}
Now, we can write 55!=55×4!=14!\dfrac{5}{5!}=\dfrac{5}{5\times 4!}=\dfrac{1}{4!} so, we will get
⇒9!4!+9!4!=10!(10−r)!\Rightarrow \dfrac{9!}{4!}+\dfrac{9!}{4!}=\dfrac{10!}{\left( 10-r \right)!}
On further simplification, we can get equation as
⇒2⋅9!4!=10!(10−r)!\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.
⇒2⋅5⋅9!5⋅4!=10!(10−r)!\Rightarrow \dfrac{2\cdot 5\cdot 9!}{5\cdot 4!}=\dfrac{10!}{\left( 10-r \right)!}
On solving we get
⇒10!5!=10!(10−r)!\Rightarrow \dfrac{10!}{5!}=\dfrac{10!}{\left( 10-r \right)!}
So, on comparing the equation we can write denominator as
⇒5=10−r\Rightarrow 5=10-r
⇒r=10−5=5\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. n!(n−r)!⋅r!\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.