Question

if (x+y)P2 = 56 and (x-y)P2 = 12 , find x and y

Answer 1

Hence,
x=6
y=2
please select as brainlist answers

Answer 2

Answer:

x = 6 and y = 2

Step-by-step explanation:

Given :  $^{(x+y)}P_2=56$ and $^{(x-y)}P_2=12$

we have to solve for x and y.

We know $^nP_r=\frac{n!}{(n-r)!}$    .....(1)

Consider $^{(x+y)}P_2=56$

apply (1) in above , we have,

n = x + y , r = 2

$^{(x+y)}P_2=\frac{(x+y)!}{(x+y-2)!}=56$

On simplify, we get, $n!=n(n-1)(n-2)...1!$ , we get,

$\frac{(x+y)(x+y-1)(x+y-2)!}{(x+y-2)!}=56$

$(x+y)(x+y-1)=56$

this means $(x+y)(x+y-1)=8\cdot 7$

Comparing both  sides we get

⇒ x + y = 8  .............(A)

Similarly,

Consider $^{(x-y)}P_2=12$

apply (1) in above , we have,

n = x - y , r = 2

$^{(x-y)}P_2=\frac{(x-y)!}{(x-y-2)!}=12$

On simplify, we get, $n!=n(n-1)(n-2)...1!$ , we get,

$\frac{(x-y)(x-y-1)(x-y-2)!}{(x-y-2)!}=12$

$(x-y)(x-y-1)=12$  

this means $(x-y)(x-y-1)=4\cdot 3$

Comparing both  sides we get

⇒ x - y = 4  ............(B)

Solving (A) and (B) , we get,

adding equation (A) and (B) , we get,

⇒ 2x = 12

⇒ x = 6

Put x = 6 in (A) , we get,

⇒ y = 2

Thus, x = 6 and y = 2