Question

Find the value of n if $^{n}{{P}_{5}}={{20}^{n}}{{P}_{3}}$.

Answer 1

Hint: Use the fact that $^{n}{{P}_{r}}=n\left( n-1 \right)\left( n-2 \right)\cdots \left( n-r+1 \right)$. Hence find the ration $\dfrac{^{n}{{P}_{5}}}{^{n}{{P}_{3}}}$ and equate it to 20. Solve the so formed quadratic equation using any of the known methods like completing the square, splitting the middle term, or using the quadratic formula. Remove the extraneous roots and hence find the value of n. Alternatively. Put n = 5, 6, … successively, till you get $^{n}{{P}_{5}}={{20}^{n}}{{P}_{3}}$. Hence find the value of n satisfying the equation.

Complete step-by-step solution -
We have $^{n}{{P}_{r}}=n\left( n-1 \right)\left( n-2 \right)\cdots \left( n-r+1 \right)$
Hence, we have
$\dfrac{^{n}{{P}_{5}}}{^{n}{{P}_{3}}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)}{n\left( n-1 \right)\left( n-2 \right)}=\left( n-3 \right)\left( n-4 \right)={{n}^{2}}-7n+12$
Hence, we have
${{n}^{2}}-7n+12=20$
Subtracting 20 from both sides, of the equation, we get
${{n}^{2}}-7n-8=0$
We have $8=8\times 1$ and $7=8-1$
Hence, we have
${{n}^{2}}-7n+8={{n}^{2}}-8n+n-8=n\left( n-8 \right)+\left( n-8 \right)=\left( n-8 \right)\left( n+1 \right)$
Hence, we have
$\left( n-8 \right)\left( n+1 \right)=0\Rightarrow n=8,-1$
Since n is a natural number, we have
$n=-1$ is rejected
Hence n =8, which is the same as obtained above.


Note: Alternative Solution
We have $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
When n = 5, we have
$\dfrac{^{5}{{P}_{5}}}{^{5}{{P}_{3}}}=\dfrac{\dfrac{5!}{0!}}{\dfrac{5!}{2!}}=2$
Hence, we have $^{n}{{P}_{5}}\ne {{20}^{n}}{{P}_{3}}$
When n = 6, we have
$\dfrac{^{6}{{P}_{5}}}{^{6}{{P}_{3}}}=\dfrac{\dfrac{6!}{1!}}{\dfrac{6!}{3!}}=6$
Hence, we have $^{n}{{P}_{5}}\ne {{20}^{n}}{{P}_{3}}$
When n = 7, we have
$\dfrac{^{7}{{P}_{5}}}{^{7}{{P}_{3}}}=\dfrac{\dfrac{7!}{2!}}{\dfrac{7!}{4!}}=12$
Hence, we have $^{n}{{P}_{5}}\ne {{20}^{n}}{{P}_{3}}$
When n =8, we have
$\dfrac{^{8}{{P}_{5}}}{^{8}{{P}_{3}}}=\dfrac{\dfrac{8!}{3!}}{\dfrac{8!}{5!}}=20$
Hence, we have $^{n}{{P}_{5}}={{20}^{n}}{{P}_{3}}$
Hence, we have n = 8.
[2] Verification:
We have $^{8}{{P}_{5}}=\dfrac{8!}{3!}=6720$ and $^{8}{{P}_{3}}=\dfrac{8!}{5!}=336$
Hence, we have $20{{\times }^{8}}{{P}_{3}}=336\times 20=6720$
Hence, we have $^{8}{{P}_{5}}=20{{\times }^{8}}{{P}_{3}}$
Hence, our answer is verified to be correct.