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Question: Find the value of n such that, (i) \({}^{n}{{P}_{5}}=42\times {}^{n}{{P}_{3}},n>4\) (ii) \(\dfra...

Find the value of n such that,
(i) nP5=42×nP3,n>4{}^{n}{{P}_{5}}=42\times {}^{n}{{P}_{3}},n>4
(ii) nP4n−1P4=53,n>4\dfrac{{}^{n}{{P}_{4}}}{{}^{n-1}{{P}_{4}}}=\dfrac{5}{3},n>4

Explanation

Solution

Hint: We will first start by using the property that nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} in part (i) and solve it to find the value of n. Then we will similarly solve the part (ii) by using the formula nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} and simplify.

Complete step-by-step answer:
Now, we will first solve the part (i) nP5=42×nP3,n>4{}^{n}{{P}_{5}}=42\times {}^{n}{{P}_{3}},n>4
Now, we know that the identity nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}.
So, using this we have,
n!(n−5)!=42×n!(n−3)!\dfrac{n!}{\left( n-5 \right)!}=42\times \dfrac{n!}{\left( n-3 \right)!}
Now, we will cross multiply the expression and cancel n! in numerator and denominator,
n!×(n−3)!(n−5)!×n!=42 (n−3)!(n−5)!=42 \begin{aligned} & \dfrac{n!\times \left( n-3 \right)!}{\left( n-5 \right)!\times n!}=42 \\\ & \dfrac{\left( n-3 \right)!}{\left( n-5 \right)!}=42 \\\ \end{aligned}
Now, we know that,
(n−3)!=(n−5)!(n−4)(n−3) =(n−5)!(n−4)(n−3)(n−5)!=42 =(n−4)(n−3)=42 \begin{aligned} & \left( n-3 \right)!=\left( n-5 \right)!\left( n-4 \right)\left( n-3 \right) \\\ & =\dfrac{\left( n-5 \right)!\left( n-4 \right)\left( n-3 \right)}{\left( n-5 \right)!}=42 \\\ & =\left( n-4 \right)\left( n-3 \right)=42 \\\ \end{aligned}
Now, we will expand the left side of the equation. So, we have,
n2−7n+12=42 n2−7n+12−42=0 n2−7n−30=0 \begin{aligned} & {{n}^{2}}-7n+12=42 \\\ & {{n}^{2}}-7n+12-42=0 \\\ & {{n}^{2}}-7n-30=0 \\\ \end{aligned}
Now, we will split the middle term as a sum of 30.
n2−10n+3n−30=0 n(n−10)+3(n−10)=0 (n+3)(n−10)=0 either n+3=0 or n−10=0 ⇒n=−3 or n=10 \begin{aligned} & {{n}^{2}}-10n+3n-30=0 \\\ & n\left( n-10 \right)+3\left( n-10 \right)=0 \\\ & \left( n+3 \right)\left( n-10 \right)=0 \\\ & either\ n+3=0\ or\ n-10=0 \\\ & \Rightarrow n=-3\ or\ n=10 \\\ \end{aligned}
Since, n > 4 has been given to us, therefore n≠−3 and n=10n\ne -3\ and\ n=10.
Now, in part (ii) we have nP4n−1P4=53,n>4\dfrac{{}^{n}{{P}_{4}}}{{}^{n-1}{{P}_{4}}}=\dfrac{5}{3},n>4.
Similarly, we will use nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. So, we have,
n!(n−4)!(n−1)!(n−5)!=53 n!×(n−5)!(n−4)!×(n−1)!=53 \begin{aligned} & \dfrac{\dfrac{n!}{\left( n-4 \right)!}}{\dfrac{\left( n-1 \right)!}{\left( n-5 \right)!}}=\dfrac{5}{3} \\\ & \dfrac{n!\times \left( n-5 \right)!}{\left( n-4 \right)!\times \left( n-1 \right)!}=\dfrac{5}{3} \\\ \end{aligned}
Now, we will write,

& n!=\left( n-1 \right)!\times n \\\ & \left( n-4 \right)!=\left( n-5 \right)!\times \left( n-4 \right) \\\ & \dfrac{\left( n-1 \right)!\times n\times \left( n-5 \right)!}{\left( n-5 \right)!\times \left( n-4 \right)\times \left( n-1 \right)!}=\dfrac{5}{3} \\\ & \dfrac{n}{n-4}=\dfrac{5}{3} \\\ \end{aligned}$$ Now, on cross – multiply we have, $\begin{aligned} & 3n=5\left( n-4 \right) \\\ & 3{{n}}-5n-20=0 \\\ & \Rightarrow 3n=5n-20 \\\ & 20=5n-3n \\\ & 2n=20 \\\ & n=10 \\\ \end{aligned}$ So, the value of n is 10 for part (ii). Note: It is important to note that while solving the first part we have used the factorization method to solve the quadratic equation and in the second part we have a linear equation to find the value of n. Also, it is important to note that we have used the fact $n!=\left( n-1 \right)n$ to solve both parts.