Question

If \[{}^{n}{{P}_{5}}=20{}^{n}{{P}_{3}}\], find the value of n.

Answer 1

Hint:The expression given 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 formula and find the value of n.

Complete step-by-step answer:
Permutation of a set is an arrangement of its arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. 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,
\[{}^{n}{{P}_{r}}=\left( n,r \right)\dfrac{n!}{\left( n-r \right)!}\]
We have been given that,
\[{}^{n}{{P}_{5}}=20{}^{n}{{P}_{3}}\]
Let us expand it with the formula told above,
\[\dfrac{n!}{\left( n-5 \right)!}=20\dfrac{n!}{\left( n-3 \right)!}\]
We can write \[n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)\left( n-5 \right)!\] for LHS and \[n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)!\] for RHS. Let us substitute it in the above expression,
\[\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)\left( n-5 \right)!}{\left( n-5 \right)!}=20\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)!}{\left( n-3 \right)!}\]
Now, cancel out the similar terms from LHS and RHS. Thus we get,
\[\begin{align}
  & \left( n-3 \right)\left( n-4 \right)=20 \\
 & {{n}^{2}}-3n-4n+12=20 \\
 & \Rightarrow {{n}^{2}}-7n+12-20=0 \\
 & \therefore {{n}^{2}}-7n-8=0 \\
\end{align}\]
The above expression is similar to the quadratic equation, \[a{{x}^{2}}+bx+c=0\]. By comparing both the expression we get,
a = 1, b = -7, c = -8
Now put these values in the quadratic formula,
\[\begin{align}
  & n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & n=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times 1\times \left( -8 \right)}}{2a}=\dfrac{7\pm \sqrt{49+32}}{2\times 1} \\
 & n=\dfrac{7\pm \sqrt{81}}{2}=\dfrac{7\pm 9}{2} \\
\end{align}\]
\[\therefore n=\dfrac{7+9}{2}=\dfrac{16}{2}=8\] or \[n=\dfrac{7-9}{2}=-1\]
The value of n cannot be negative. So, we take n = 8.
\[\therefore \] We got the value of n = 8.

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 thing taken r at a time,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].We can also verify the answer by substituting value n=8 in the expression \[{}^{n}{{P}_{5}}=20{}^{n}{{P}_{3}}\] and check whether L.H.S=R.H.S or not.