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Question: The equation is \({}^{n}{{p}_{4}}=20\times {}^{n}{{p}_{2}}\). Find the value of n....

The equation is np4=20×np2{}^{n}{{p}_{4}}=20\times {}^{n}{{p}_{2}}. Find the value of n.

Explanation

Solution

We first define the notion of factorial and permutation. We express them in their mathematical form and find the expansion. Then we solve the given equation to find a quadratic expression of n. we solve the equation for the solution of the problem.

Complete step-by-step answer:
We first define the notion of two things which are factorial and permutation.
By using the term factorial of any particular natural number, we mean the multiplication of all the numbers starting from 1 to that particular number.
So, mathematical expression is n!=1×2×3×4×.......×(n−1)×nn!=1\times 2\times 3\times 4\times .......\times \left( n-1 \right)\times n. Here n∈Nn\in \mathbb{N}.
For example: 5!=1×2×3×4×5=1205!=1\times 2\times 3\times 4\times 5=120.
For the permutation part, by using the term npr{}^{n}{{p}_{r}}, we mean the arrangement of r terms after choosing them out of n terms.
So, mathematical expression is npr=n!(n−r)!{}^{n}{{p}_{r}}=\dfrac{n!}{\left( n-r \right)!}. Here n≥r≥0,n≠0n\ge r\ge 0,n\ne 0.
For example: 5p3=5!(5−3)!=5!2!=1202=60{}^{5}{{p}_{3}}=\dfrac{5!}{\left( 5-3 \right)!}=\dfrac{5!}{2!}=\dfrac{120}{2}=60.
Now we try to solve the given equation np4=20×np2{}^{n}{{p}_{4}}=20\times {}^{n}{{p}_{2}}.
np4=20×np2 ⇒n!(n−4)!=20×n!(n−2)! \begin{aligned} & {}^{n}{{p}_{4}}=20\times {}^{n}{{p}_{2}} \\\ & \Rightarrow \dfrac{n!}{\left( n-4 \right)!}=20\times \dfrac{n!}{\left( n-2 \right)!} \\\ \end{aligned}
We take all unknowns in one side and get
n!×(n−2)!(n−4)!×n!=20 ⇒(n−2)(n−3)(n−4)!(n−4)!=20 ⇒(n−2)(n−3)=20 \begin{aligned} & \dfrac{n!\times \left( n-2 \right)!}{\left( n-4 \right)!\times n!}=20 \\\ & \Rightarrow \dfrac{\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)!}{\left( n-4 \right)!}=20 \\\ & \Rightarrow \left( n-2 \right)\left( n-3 \right)=20 \\\ \end{aligned}
We got a quadratic equation of n and now we solve it to find the value of n.
(n−2)(n−3)=20 ⇒n2−5n+6=20 ⇒n2−5n−14=0 ⇒n2−7n+2n−14=0 ⇒(n−7)(n+2)=0 \begin{aligned} & \left( n-2 \right)\left( n-3 \right)=20 \\\ & \Rightarrow {{n}^{2}}-5n+6=20 \\\ & \Rightarrow {{n}^{2}}-5n-14=0 \\\ & \Rightarrow {{n}^{2}}-7n+2n-14=0 \\\ & \Rightarrow \left( n-7 \right)\left( n+2 \right)=0 \\\ \end{aligned}
From the multiplication form we get 2 solutions of n. so, n=−2,7n=-2,7.
We already mentioned that npr=n!(n−r)!{}^{n}{{p}_{r}}=\dfrac{n!}{\left( n-r \right)!}. Here n≥r≥0,n≠0n\ge r\ge 0,n\ne 0.
So, n can’t be -2. Value of n will be 7.

Note: We always need to remember to cross-check the solution of the problem with the main equation to validate. Just like the given problem all the solutions need not to satisfy the equation as the equation is not made up with the conditions. Also, the given problem involves only permutation. In case of combination the formula would be different.