Solveeit Logo
Login

Question

Question: Find the value of n if \(^{n}{{P}_{5}}={{20}^{n}}{{P}_{3}}\)....

Find the value of n if nP5=20nP3^{n}{{P}_{5}}={{20}^{n}}{{P}_{3}}.

Explanation

Solution

Hint: Use the fact that nPr=n(n−1)(n−2)⋯(n−r+1)^{n}{{P}_{r}}=n\left( n-1 \right)\left( n-2 \right)\cdots \left( n-r+1 \right). Hence find the ration nP5nP3\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 nP5=20nP3^{n}{{P}_{5}}={{20}^{n}}{{P}_{3}}. Hence find the value of n satisfying the equation.

Complete step-by-step solution -
We have nPr=n(n−1)(n−2)⋯(n−r+1)^{n}{{P}_{r}}=n\left( n-1 \right)\left( n-2 \right)\cdots \left( n-r+1 \right)
Hence, we have
nP5nP3=n(n−1)(n−2)(n−3)(n−4)n(n−1)(n−2)=(n−3)(n−4)=n2−7n+12\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
n2−7n+12=20{{n}^{2}}-7n+12=20
Subtracting 20 from both sides, of the equation, we get
n2−7n−8=0{{n}^{2}}-7n-8=0
We have 8=8×18=8\times 1 and 7=8−17=8-1
Hence, we have
n2−7n+8=n2−8n+n−8=n(n−8)+(n−8)=(n−8)(n+1){{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
(n−8)(n+1)=0⇒n=8,−1\left( n-8 \right)\left( n+1 \right)=0\Rightarrow n=8,-1
Since n is a natural number, we have
n=−1n=-1 is rejected
Hence n =8, which is the same as obtained above.

Note: Alternative Solution
We have nPr=n!(n−r)!^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}
When n = 5, we have
5P55P3=5!0!5!2!=2\dfrac{^{5}{{P}_{5}}}{^{5}{{P}_{3}}}=\dfrac{\dfrac{5!}{0!}}{\dfrac{5!}{2!}}=2
Hence, we have nP5≠20nP3^{n}{{P}_{5}}\ne {{20}^{n}}{{P}_{3}}
When n = 6, we have
6P56P3=6!1!6!3!=6\dfrac{^{6}{{P}_{5}}}{^{6}{{P}_{3}}}=\dfrac{\dfrac{6!}{1!}}{\dfrac{6!}{3!}}=6
Hence, we have nP5≠20nP3^{n}{{P}_{5}}\ne {{20}^{n}}{{P}_{3}}
When n = 7, we have
7P57P3=7!2!7!4!=12\dfrac{^{7}{{P}_{5}}}{^{7}{{P}_{3}}}=\dfrac{\dfrac{7!}{2!}}{\dfrac{7!}{4!}}=12
Hence, we have nP5≠20nP3^{n}{{P}_{5}}\ne {{20}^{n}}{{P}_{3}}
When n =8, we have
8P58P3=8!3!8!5!=20\dfrac{^{8}{{P}_{5}}}{^{8}{{P}_{3}}}=\dfrac{\dfrac{8!}{3!}}{\dfrac{8!}{5!}}=20
Hence, we have nP5=20nP3^{n}{{P}_{5}}={{20}^{n}}{{P}_{3}}
Hence, we have n = 8.
[2] Verification:
We have 8P5=8!3!=6720^{8}{{P}_{5}}=\dfrac{8!}{3!}=6720 and 8P3=8!5!=336^{8}{{P}_{3}}=\dfrac{8!}{5!}=336
Hence, we have 20×8P3=336×20=672020{{\times }^{8}}{{P}_{3}}=336\times 20=6720
Hence, we have 8P5=20×8P3^{8}{{P}_{5}}=20{{\times }^{8}}{{P}_{3}}
Hence, our answer is verified to be correct.