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Question: If\[{}^{2n - 1}{P_n}:{}^{2n + 1}{P_{n - 1}} = 22:7\] , then find the value of n....

If2n−1Pn:2n+1Pn−1=22:7{}^{2n - 1}{P_n}:{}^{2n + 1}{P_{n - 1}} = 22:7 , then find the value of n.

Explanation

Solution

We know that the number of permutations of n things taken k at a time is denoted by nPk{}^n{P_k}and defined by,
nPk=n!(n−k)!{}^n{P_k} = \dfrac{{n!}}{{(n - k)!}}
Where, n! is factorial of n and defined by
n!=n(n−1)(n−2).......2.1n! = n(n - 1)(n - 2).......2.1

Complete step-by-step answer:
It is given that, 2n−1Pn:2n+1Pn−1=22:7{}^{2n - 1}{P_n}:{}^{2n + 1}{P_{n - 1}} = 22:7
Now using the definition of ratios we get
2n−1Pn:2n+1Pn−1=2n−1Pn2n+1Pn−1{}^{2n - 1}{P_n}:{}^{2n + 1}{P_{n - 1}} = \dfrac{{{}^{2n - 1}{P_n}}}{{{}^{2n + 1}{P_{n - 1}}}}
Now let us apply the permutation formula so that we could arrive at the following condition,
2n−1Pn2n+1Pn−1=(2n−1)!(2n−1−n)!(2n+1)!(2n+1−n+1)!\dfrac{{{}^{2n - 1}{P_n}}}{{{}^{2n + 1}{P_{n - 1}}}} = \dfrac{{\dfrac{{(2n - 1)!}}{{(2n - 1 - n)!}}}}{{\dfrac{{(2n + 1)!}}{{(2n + 1 - n + 1)!}}}}
Now let us simply the values, then we get,
(2n−1)!(2n−1−n)!(2n+1)!(2n+1−n+1)!=(2n−1)!(n−1)!(2n+1)!(n+2)!\dfrac{{\dfrac{{(2n - 1)!}}{{(2n - 1 - n)!}}}}{{\dfrac{{(2n + 1)!}}{{(2n + 1 - n + 1)!}}}} = \dfrac{{\dfrac{{(2n - 1)!}}{{(n - 1)!}}}}{{\dfrac{{(2n + 1)!}}{{(n + 2)!}}}}
On solving the above equation and applying the factorial value, we get

(2n−1)!(n−1)!(2n+1)!(n+2)!=(2n−1)!(n−1)!×(n+2)!(2n+1)! =(2n−1)!(n−1)!×(n+2)(n+1)n(n−1)!(2n+1)2n(2n−1)!  \dfrac{{\dfrac{{(2n - 1)!}}{{(n - 1)!}}}}{{\dfrac{{(2n + 1)!}}{{(n + 2)!}}}} = \dfrac{{(2n - 1)!}}{{(n - 1)!}} \times \dfrac{{(n + 2)!}}{{(2n + 1)!}} \\\ = \dfrac{{(2n - 1)!}}{{(n - 1)!}} \times \dfrac{{(n + 2)(n + 1)n(n - 1)!}}{{(2n + 1)2n(2n - 1)!}} \\\

By cancelling common terms from numerator & denominator most of the factorial terms get cancelled, then we get

=(n+2)(n+1)n(2n+1)2n =(n+2)(n+1)(4n+2)  = \dfrac{{(n + 2)(n + 1)n}}{{(2n + 1)2n}} \\\ = \dfrac{{(n + 2)(n + 1)}}{{(4n + 2)}} \\\

Thus we are also provided that2n−1Pn:2n+1Pn−1=22:7{}^{2n - 1}{P_n}:{}^{2n + 1}{P_{n - 1}} = 22:7 comparing this with the above final expression we have,
(n+2)(n+1)(4n+2)=227\dfrac{{(n + 2)(n + 1)}}{{(4n + 2)}} = \dfrac{{22}}{7}
Let us solve this equation, so that we get,

7(n+2)(n+1)=22(4n+2) 7(n2+n+2n+2)=88n+44  7(n + 2)(n + 1) = 22(4n + 2) \\\ 7({n^2} + n + 2n + 2) = 88n + 44 \\\

On further reducing the above equation we get,

7(n2+3n+2)=88n+44 7n2+21n+14−88n−44=0  7({n^2} + 3n + 2) = 88n + 44 \\\ 7{n^2} + 21n + 14 - 88n - 44 = 0 \\\

Thus on solving it we arrive at the following quadratic equation,
7n2−67n−30=07{n^2} - 67n - 30 = 0
Let us now solve the quadratic equation to find the value of n,

7n2−70n+3n−30=0 7n(n−10)+3(n−10)=0  7{n^2} - 70n + 3n - 30 = 0 \\\ 7n(n - 10) + 3(n - 10) = 0 \\\

On solving it again, we get

(n−10)(7n+3)=0 (n−10)=0 or (7n+3)=0  (n - 10)(7n + 3) = 0 \\\ (n - 10) = 0{\text{ or }}(7n + 3) = 0 \\\

n=10 or,−37or, - \dfrac{3}{7}

Hence we have found that the value of n is either 10 or −37 - \dfrac{3}{7}

Note:
Quadratic equations can be solved by a middle term process.
Splitting the middle term either addition of two numbers or subtraction of two numbers, then make the quadratic equation as the multiplication of two factors.
Solving the two factors of the quadratic equation we will get the solution.
The roots of the quadratic equation can also be found by using the sridhar acharya's formula.