Solveeit Logo
Login

Question

Question: If \[{}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5\], then find the value of n....

If 2n+1Pn−1:2n−1Pn=3:5{}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5, then find the value of n.

Explanation

Solution

Hint:The expression 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 value of n.

Complete step-by-step answer:
Permutation of a set is an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its element. 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=P(n,r)n!(n−r)!{}^{n}{{P}_{r}}=P\left( n,r \right)\dfrac{n!}{\left( n-r \right)!}
Now, we have been given that,
2n+1Pn−1:2n−1Pn=3:5…...(1){}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5 …...(1)
2n+1Pn−1=(2n+1)!((2n+1)−(n−1))!{}^{2n+1}{{P}_{n-1}}=\dfrac{\left( 2n+1 \right)!}{\left( \left( 2n+1 \right)-\left( n-1 \right) \right)!}
Let us simplify the above expression for 2n+1Pn−1{}^{2n+1}{{P}_{n-1}}.
2n+1Pn−1=(2n+1)!(2n+1−n+1)!=(2n+1)!(n+2)!{}^{2n+1}{{P}_{n-1}}=\dfrac{\left( 2n+1 \right)!}{\left( 2n+1-n+1 \right)!}=\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}
∴2n+1Pn−1=(2n+1)!(n+2)!........(2)\therefore {}^{2n+1}{{P}_{n-1}}=\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}........(2)
Similarly, 2n−1Pn=(2n+1)!(2n−1−n)!=(2n−1)!(n−1)!.........(3){}^{2n-1}{{P}_{n}}=\dfrac{\left( 2n+1 \right)!}{\left( 2n-1-n \right)!}=\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!}.........(3)
Now let us put the value of (2) and (3) in (1). We get,
2n+1Pn−1:2n−1Pn=2n+1Pn−12n−1Pn=(2n+1)!(n+2)!(2n−1)!(n−1)!{}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=\dfrac{{}^{2n+1}{{P}_{n-1}}}{{}^{2n-1}{{P}_{n}}}=\dfrac{\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}}{\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!}}
Now let us simplify the above expression.
Thus we can write it as,

& \dfrac{\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}}{\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!}}=\dfrac{\left( 2n+1 \right)!}{\left( n+2 \right)!}\times \dfrac{\left( n-1 \right)!}{\left( 2n-1 \right)!} \\\ & \left( 2n+1 \right)!=\left( 2n+1 \right)\times 2n\times \left( 2n-1 \right)! \\\ \end{aligned}$$ Similarly, $$\left( n+2 \right)!=\left( n+2 \right)\left( n+1 \right)n\left( n-1 \right)!$$. Substitute these in the above and simplify it, $$\dfrac{\left( 2n+1 \right)2n\left( 2n-1 \right)!}{\left( n+2 \right)\left( n+1 \right)n\left( n-1 \right)!}\times \dfrac{\left( n-1 \right)!}{\left( 2n-1 \right)!}=\dfrac{\left( 2n+1 \right)\times 2n}{\left( n+2 \right)\left( n+1 \right)n}$$ Now, we know that $${}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5$$. Thus we can write the above expression as, $$\dfrac{2\left( 2n+1 \right)}{\left( n+2 \right)\left( n+1 \right)}=\dfrac{3}{5}$$, Apply cross multiplication property and simplify it, $$\begin{aligned} & 10\left( 2n+1 \right)=3\left( n+1 \right)\left( n+2 \right) \\\ & 20n+10=3\left[ {{n}^{2}}+2n+n+2 \right] \\\ & \Rightarrow 20n+10=3{{n}^{2}}+9n+6 \\\ & \therefore 3{{n}^{2}}+9n-20n+6-10=0 \\\ & \Rightarrow 3{{n}^{2}}-11n-4=0 \\\ \end{aligned}$$ The above expression is similar to the quadratic equation $$a{{x}^{2}}+bx+c=0$$. By comparing both the equation, we can say that, a = 3, b = -11, c = -4 Now put these values in the quadratic formula, $$n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ $$\begin{aligned} & n=\dfrac{-\left( -11 \right)\pm \sqrt{{{\left( -11 \right)}^{2}}-4\times 3\times \left( -4 \right)}}{2\times 3}=\dfrac{11\pm \sqrt{121+48}}{6} \\\ & n=\dfrac{11\pm \sqrt{169}}{6}=\dfrac{11\pm 13}{6} \\\ \end{aligned}$$ Thus we get the value of $$n=\dfrac{11+13}{6}$$ and $$n=\dfrac{11-13}{6}$$. i.e. $$n=\dfrac{24}{6}=4$$ and $$n=\dfrac{-1}{3}$$. While solving permutation, the values can’t be negative. Thus the only possible solution is that, n = 4. Note: Don’t confuse the formula of permutation with the formula of combination. In combination, the order does not matter. Thus the number of combinations will have the formula, n things taken r at a time: $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$.We can also verify the answer by substituting n=4 in the expression $${}^{2n+1}{{P}_{n-1}}:{}^{2n-1}{{P}_{n}}=3:5$$.