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Question: If \[{}^{n}{{P}_{5}}=20{}^{n}{{P}_{3}}\], find the value of n....

If nP5=20nP3{}^{n}{{P}_{5}}=20{}^{n}{{P}_{3}}, find the value of n.

Explanation

Solution

Hint:The expression given is that of Permutation, which represents ordered matters. For number of permutation of n things taken r at a time = nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. Simplify the given expression with this formula and find the formula and find the value of n.

Complete step-by-step answer:
Permutation of a set is an arrangement of its arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. Permutation is also the linear order of an ordered set. Thus the number of permutation (ordered matters) of n things taken r at a time is given as,
nPr=(n,r)n!(n−r)!{}^{n}{{P}_{r}}=\left( n,r \right)\dfrac{n!}{\left( n-r \right)!}
We have been given that,
nP5=20nP3{}^{n}{{P}_{5}}=20{}^{n}{{P}_{3}}
Let us expand it with the formula told above,
n!(n−5)!=20n!(n−3)!\dfrac{n!}{\left( n-5 \right)!}=20\dfrac{n!}{\left( n-3 \right)!}
We can write n!=n(n−1)(n−2)(n−3)(n−4)(n−5)!n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)\left( n-5 \right)! for LHS and n!=n(n−1)(n−2)(n−3)!n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)! for RHS. Let us substitute it in the above expression,
n(n−1)(n−2)(n−3)(n−4)(n−5)!(n−5)!=20n(n−1)(n−2)(n−3)!(n−3)!\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\left( n-4 \right)\left( n-5 \right)!}{\left( n-5 \right)!}=20\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)!}{\left( n-3 \right)!}
Now, cancel out the similar terms from LHS and RHS. Thus we get,

& \left( n-3 \right)\left( n-4 \right)=20 \\\ & {{n}^{2}}-3n-4n+12=20 \\\ & \Rightarrow {{n}^{2}}-7n+12-20=0 \\\ & \therefore {{n}^{2}}-7n-8=0 \\\ \end{aligned}$$ The above expression is similar to the quadratic equation, $$a{{x}^{2}}+bx+c=0$$. By comparing both the expression we get, a = 1, b = -7, c = -8 Now put these values in the quadratic formula, $$\begin{aligned} & n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\\ & n=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times 1\times \left( -8 \right)}}{2a}=\dfrac{7\pm \sqrt{49+32}}{2\times 1} \\\ & n=\dfrac{7\pm \sqrt{81}}{2}=\dfrac{7\pm 9}{2} \\\ \end{aligned}$$ $$\therefore n=\dfrac{7+9}{2}=\dfrac{16}{2}=8$$ or $$n=\dfrac{7-9}{2}=-1$$ The value of n cannot be negative. So, we take n = 8. $$\therefore $$ We got the value of n = 8. Note: Don’t confuse the formula of Permutation with the formula of combination. In combination, the order does not matter. Thus the number of combinations will have the formula, n thing taken r at a time, $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$.We can also verify the answer by substituting value n=8 in the expression $${}^{n}{{P}_{5}}=20{}^{n}{{P}_{3}}$$ and check whether L.H.S=R.H.S or not.