Question

If ${}^{{\text{m + n}}}{{\text{P}}_2} = 90{\text{ and }}{}^{{\text{m - n}}}{{\text{P}}_2}$ =30, then (m, n) =
A. (8, 2)
B. (9, 2)
C. (8, 1)
D. (7, 3)

Answer 1

Hint: The strategy for solving the question is that first write the formula to find the permutation of m+n things taken 2 at a time to get an equation in m and n. Do similarly with another given expression to get one more equation. Solve these two equations to get the values of m and n.

Complete step-by-step answer:

In the question, it is given that:
${}^{{\text{m + n}}}{{\text{P}}_2} = 90{\text{ and }}{}^{{\text{m - n}}}{{\text{P}}_2}$=30 and we have to find the value of m and n.

We know that the permutation of n things taken r at a time is calculated as follow:
${}^{\text{n}}{{\text{P}}_r} = \dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right)!}}$
$\therefore $ ${}^{{\text{m + n}}}{{\text{P}}_2}$ =$\dfrac{{{\text{(m + n)!}}}}{{\left( {{\text{m + n - 2}}} \right)!}}$ and ${}^{{\text{m - n}}}{{\text{P}}_2}$=$\dfrac{{{\text{(m - n)!}}}}{{\left( {{\text{m - n - 2}}} \right)!}}$.

On further solving the above equation, we get:
${}^{{\text{m + n}}}{{\text{P}}_2}$=$\dfrac{{{\text{(m + n)!}}}}{{\left( {{\text{m + n - 2}}} \right)!}}$=$\dfrac{{({\text{m + n)(m + n - 1)(m + n - 2)(m + n - 3)}} \times ... \times {\text{1}}}}{{({\text{m + n - 2)(m + n - 3)}} \times ... \times {\text{1}}}}$ =(m+n)(m+n-1).
${}^{{\text{m - n}}}{{\text{P}}_2}$=$\dfrac{{{\text{(m - n)!}}}}{{\left( {{\text{m - n - 2}}} \right)!}}$=$\dfrac{{({\text{m - n)(m - n - 1)(m - n - 2)(m - n - 3)}} \times ... \times {\text{1}}}}{{({\text{m - n - 2)(m - n - 3)}} \times ... \times {\text{1}}}}$ =(m-n)(m-n-1).
According to the question:
${}^{{\text{m + n}}}{{\text{P}}_2} = 90$
$ \Rightarrow $ (m+n)(m+n-1) = 90 $\to$ (1)

Let us assume that (m+n) =x.
Equation 1 can be written as:
x(x-1) = 90
$ \Rightarrow {{\text{x}}^2}{\text{ - x - 90 = 0}}$
$ \Rightarrow $ (x-10)(x+9)=0
x =10 and x = - 9.

But (m+n) cannot be negative.
$\therefore $ m+n = 10 $\to$ (2)

Also
${}^{{\text{m - n}}}{{\text{P}}_2}$ = 30
$ \Rightarrow $ (m-n)(m-n-1) = 30

Let us assume that (m-n) =y.

Equation 1 can be written as:
y(y-1) = 30
$ \Rightarrow {{\text{y}}^2}{\text{ - y - 30 = 0}}$
$ \Rightarrow $ (y-6)(y+5) = 0
y = 6 and y = -5

But (m-n) cannot be negative.
$\therefore $ m-n = 6 $\to$ (3)

On adding the equation 2 and 3, we get:
2m = 16
$ \Rightarrow $ m = 8

Putting the values in equation 2, we get:
m+n = 10
$ \Rightarrow $ 8+n= 10
$ \Rightarrow $ n = 2.

Therefore, (m,n) is (8, 2).

So option A is correct.

Note: In this type of question first you should know how to find the permutation of n things taken r at a time. You should know that n! = n(n-1)(n-2)(n-3)$ \times $ ....2$ \times $ 1.Do not just multiply the terms of the equation otherwise you will get an equation with degree 4 which will take a lot of time to solve.