Question

If \[{}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5\], then find the value of n.

Answer 1

Hint:The expression is that of Permutation, which represents ordered matters. For number of permutation of n things taken r at a time = \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]. Simplify the given expression with this formula and find the value of n.

Complete step-by-step answer:
Permutation of a set is an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its element. Permutation is also the linear order of an ordered set. Thus the number of permutation (ordered matters) of n things taken r at a time is given as,
\[{}^{n}{{P}_{r}}=P\left( n,r \right)\dfrac{n!}{\left( n-r \right)!}\]
Now, we have been given that,
\[{}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5 …...(1) \]
\[{}^{2n+1}{{P}_{n-1}}=\dfrac{\left( 2n+1 \right)!}{\left( \left( 2n+1 \right)-\left( n-1 \right) \right)!}\]
Let us simplify the above expression for \[{}^{2n+1}{{P}_{n-1}}\].
\[{}^{2n+1}{{P}_{n-1}}=\dfrac{\left( 2n+1 \right)!}{\left( 2n+1-n+1 \right)!}=\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}\]
\[\therefore {}^{2n+1}{{P}_{n-1}}=\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}........(2)\]
Similarly, \[{}^{2n-1}{{P}_{n}}=\dfrac{\left( 2n+1 \right)!}{\left( 2n-1-n \right)!}=\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!}.........(3)\]
Now let us put the value of (2) and (3) in (1). We get,
\[{}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=\dfrac{{}^{2n+1}{{P}_{n-1}}}{{}^{2n-1}{{P}_{n}}}=\dfrac{\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}}{\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!}}\]
Now let us simplify the above expression.
Thus we can write it as,
\[\begin{align}
  & \dfrac{\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}}{\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!}}=\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}\times \dfrac{\left( n-1 \right)!}{\left( 2n-1 \right)!} \\
 & \left( 2n+1 \right)!=\left( 2n+1 \right)\times 2n\times \left( 2n-1 \right)! \\
\end{align}\]
Similarly, \[\left( n+2 \right)!=\left( n+2 \right)\left( n+1 \right)n\left( n-1 \right)!\].
Substitute these in the above and simplify it,
\[\dfrac{\left( 2n+1 \right)2n\left( 2n-1 \right)!}{\left( n+2 \right)\left( n+1 \right)n\left( n-1 \right)!}\times \dfrac{\left( n-1 \right)!}{\left( 2n-1 \right)!}=\dfrac{\left( 2n+1 \right)\times 2n}{\left( n+2 \right)\left( n+1 \right)n}\]
Now, we know that \[{}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5\]. Thus we can write the above expression as,
\[\dfrac{2\left( 2n+1 \right)}{\left( n+2 \right)\left( n+1 \right)}=\dfrac{3}{5}\], Apply cross multiplication property and simplify it,
\[\begin{align}
  & 10\left( 2n+1 \right)=3\left( n+1 \right)\left( n+2 \right) \\
 & 20n+10=3\left[ {{n}^{2}}+2n+n+2 \right] \\
 & \Rightarrow 20n+10=3{{n}^{2}}+9n+6 \\
 & \therefore 3{{n}^{2}}+9n-20n+6-10=0 \\
 & \Rightarrow 3{{n}^{2}}-11n-4=0 \\
\end{align}\]
The above expression is similar to the quadratic equation \[a{{x}^{2}}+bx+c=0\]. By comparing both the equation, we can say that,
a = 3, b = -11, c = -4
Now put these values in the quadratic formula, \[n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
\[\begin{align}
  & n=\dfrac{-\left( -11 \right)\pm \sqrt{{{\left( -11 \right)}^{2}}-4\times 3\times \left( -4 \right)}}{2\times 3}=\dfrac{11\pm \sqrt{121+48}}{6} \\
 & n=\dfrac{11\pm \sqrt{169}}{6}=\dfrac{11\pm 13}{6} \\
\end{align}\]
Thus we get the value of \[n=\dfrac{11+13}{6}\] and \[n=\dfrac{11-13}{6}\].
i.e. \[n=\dfrac{24}{6}=4\] and \[n=\dfrac{-1}{3}\].
While solving permutation, the values can’t be negative. Thus the only possible solution is that, n = 4.

Note: Don’t confuse the formula of permutation with the formula of combination. In combination, the order does not matter. Thus the number of combinations will have the formula, n things taken r at a time:
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].We can also verify the answer by substituting n=4 in the expression \[{}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5\].