Explanation:

\(^{m+n}P_4 = 3024\)

\(\frac{(m+n)!}{(m+n-4)!} = 3024\)

\(\frac{(m+n)(m+n-1)(m+n-2)(m+n-3)(m+n-4)!}{(m+n-4)!} = 3024\)

\((m+n)(m+n-1)(m+n-2)(m+n-3) = 3024\)

For easy calculation, let \((m+n) = x\)

So, \(x(x-1)(x-2)(x-3) = 3024\)

Now, we will factorize 3024 and try to make it as a product of 4 consecutive numbers = \((9)(8)(7)(6)\)
Therefore, \(x = m+n = 9\)

Similarly,
\(^{m-n}P_4 = 120\)

\(\frac{(m-n)!}{(m-n-4)!} = 120\)

let \(m-n = y\)
So, \(y(y-1)(y-2)(y-3) = 120 = (5)(4)(3)(2)\)

Therefore, \(y = m-n = 5\)

Solve two linear equations to get \(m = 7\) and \(n = 2\)

Col. A: \((7)(2)\)
Col. B: \(14\)

Hence, option C