Solveeit Logo
Login

Question

Question: If \({}^{{\text{m + n}}}{{\text{P}}_2} = 90{\text{ and }}{}^{{\text{m - n}}}{{\text{P}}_2}\) =30, th...

If m + nP2=90 and m - nP2{}^{{\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)

Explanation

Solution

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:
m + nP2=90 and m - nP2{}^{{\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:
nPr=n!(n - r)!{}^{\text{n}}{{\text{P}}_r} = \dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right)!}}
∴\therefore m + nP2{}^{{\text{m + n}}}{{\text{P}}_2} =(m + n)!(m + n - 2)!\dfrac{{{\text{(m + n)!}}}}{{\left( {{\text{m + n - 2}}} \right)!}} and m - nP2{}^{{\text{m - n}}}{{\text{P}}_2}=(m - n)!(m - n - 2)!\dfrac{{{\text{(m - n)!}}}}{{\left( {{\text{m - n - 2}}} \right)!}}.

On further solving the above equation, we get:
m + nP2{}^{{\text{m + n}}}{{\text{P}}_2}=(m + n)!(m + n - 2)!\dfrac{{{\text{(m + n)!}}}}{{\left( {{\text{m + n - 2}}} \right)!}}=(m + n)(m + n - 1)(m + n - 2)(m + n - 3)×...×1(m + n - 2)(m + n - 3)×...×1\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).
m - nP2{}^{{\text{m - n}}}{{\text{P}}_2}=(m - n)!(m - n - 2)!\dfrac{{{\text{(m - n)!}}}}{{\left( {{\text{m - n - 2}}} \right)!}}=(m - n)(m - n - 1)(m - n - 2)(m - n - 3)×...×1(m - n - 2)(m - n - 3)×...×1\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:
m + nP2=90{}^{{\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
⇒x2 - x - 90 = 0\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
m - nP2{}^{{\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
⇒y2 - y - 30 = 0\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.