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Question: Find value of n if \({}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right)\)....

Find value of n if nP4=5(nP3){}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right).

Explanation

Solution

Hint: We know that permutation of nPr{}^{n}{{P}_{r}} can be given, by using the formula, nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. So, by comparing it with nP4{}^{n}{{P}_{4}} and nP3{}^{n}{{P}_{3}}, then equating them with each other and using the formula, (n−r)!=(n−r)(n−(r+1))!\left( n-r \right)!=\left( n-r \right)\left( n-\left( r+1 \right) \right)! wherever needed, we will find the value of n.

Complete step-by-step solution -
In question we are given that nP4=5(nP3){}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right) and we are asked to find the value of n, so first of all we will compare it with general expression i.e. nPr{}^{n}{{P}_{r}} and we know that its value in form of permutation can be given by,
nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} ……………(i)
Now, on comparing nP4{}^{n}{{P}_{4}} with expression (i) and we can say that for nP4{}^{n}{{P}_{4}}, r=4r=4 and n=nn=n, so now we can write nP4{}^{n}{{P}_{4}} as,
nP4=n!(n−4)!{}^{n}{{P}_{4}}=\dfrac{n!}{\left( n-4 \right)!} ………….(ii)
In the same way, on comparing nP3{}^{n}{{P}_{3}} with nPr{}^{n}{{P}_{r}}, we can say that for nP3{}^{n}{{P}_{3}}, r=3r=3 and n=nn=n, so now, we can write nP3{}^{n}{{P}_{3}} as,
nP4=n!(n−3)!{}^{n}{{P}_{4}}=\dfrac{n!}{\left( n-3 \right)!} ……………….(iii)
Now, in question it is given that nP4=5(nP3){}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right), so on substituting the value of nP4{}^{n}{{P}_{4}} and nP3{}^{n}{{P}_{3}} from expression (ii) and (iii) we will get,
n!(n−4)!=5(n!(n−3)!)\dfrac{n!}{\left( n-4 \right)!}=5\left( \dfrac{n!}{\left( n-3 \right)!} \right)
On further simplifying we will get,
⇒n!(n−4)!=5n!(n−3)!\Rightarrow \dfrac{n!}{\left( n-4 \right)!}=\dfrac{5n!}{\left( n-3 \right)!}
Now, the expansion of (n−3)!\left( n-3 \right)!, can be given as, (n−3)!=(n−3)(n−4)!\left( n-3 \right)!=\left( n-3 \right)\left( n-4 \right)!, by using the formula (n−r)!=(n−r)(n−(r+1))!\left( n-r \right)!=\left( n-r \right)\left( n-\left( r+1 \right) \right)!, so on again substituting this value in expression and canceling n!n! in numerator as they are same, we will get,
⇒1(n−4)!=5(n−3)(n−4)!\Rightarrow \dfrac{1}{\left( n-4 \right)!}=\dfrac{5}{\left( n-3 \right)\left( n-4 \right)!}
Now, on cancelling similar terms i.e. (n−4)!\left( n-4 \right)!, we will get,
⇒1=5n−3\Rightarrow 1=\dfrac{5}{n-3}
⇒1(n−3)=5\Rightarrow 1\left( n-3 \right)=5
⇒n−3=5\Rightarrow n-3=5
⇒n=5+3=8\Rightarrow n=5+3=8
Thus, we can say that value of n in nP4=5(nP3){}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right) is 8.

Note: Instead of applying the formula, (n−r)!=(n−r)(n−(r+1))!\left( n-r \right)!=\left( n-r \right)\left( n-\left( r+1 \right) \right)!, in (n−3)!\left( n-3 \right)!, students might apply in (n−4)!\left( n-4 \right)!, such as (n−4)!=(n−4)(n−(4+1))!\left( n-4 \right)!=\left( n-4 \right)\left( n-\left( 4+1 \right) \right)!, which will be wrong and students won't be able to simplify the equation. So, students must use formulas properly and avoid the mistakes to get answers.