Question

If ${}^{m+n}{P}_2=90$ and ${}^{m-n}{P}_2=30$, then (m, n) =

• A. (8,2)
• B. (9, 2)
• C. (8, 1)
• D. (7, 3)

Answer 1

The correct option is A (8,2)

${}^{m+n}{P}_2=90$

$\Rightarrow \frac{(m+n)!}{(m+n-2)!}=90$

$(m+n)(m+n-1)=90$

Let $m+n=p$

$\Rightarrow p\times p-1=90$

$\Rightarrow p\times p-1=10\times 9$

$\Rightarrow p=10$

${}^{m-n}{P}_2=30$

$\Rightarrow \frac{(m-n)!}{(m-n-2)!}=90$

$(m-n)(m-n-1)=30$

Let $m-n=q$

$\Rightarrow q\times q-1=90$

$\Rightarrow q\times q-1=6\times 5$

$\Rightarrow q=6$

$m+n=10$

$m-n=6$

$2m=16\Rightarrow m=8,n=2$

$\therefore (m,n)=(8,2)$