Question

If $\rm ^{(x+y)}P_{2} = 56$ and $\rm ^{(x-y)}P_{2} = 12$, find x and y.

Answer 1

We know that

$\rm ^{n}P_{r} = \frac{n!}{(n - r)!} \\ \\ \rm ^{(x+y)}P_{2} = 56 \\ \\ \frac{(x + y)!}{(x + y - 2)!} = 56 \\ \\ \frac{(x + y)(x + y - 1)(x + y - 2)!}{(x + y - 2)!} = 56 \\ \\ (x + y)(x + y - 1) = 56 \: \: \: \: \\ \\ {x}^{2} + xy - x + xy + {y}^{2} - y = 56 \\ \\ {x}^{2} + {y}^{2} + 2xy - x - y = 56...eq1$

$\rm ^{(x-y)}P_{2} = 12 \\ \\ \frac{(x - y)!}{(x - y - 2)!} = 12 \\ \\ \frac{(x - y)(x - y - 1)(x - y - 2)!}{(x - y - 2)!} = 12 \\ \\ (x - y)(x - y - 1) = 12 \: \: \: \: \\ \\ {x}^{2} - xy - x - xy + {y}^{2} + y = 12 \\ \\ {x}^{2} + {y}^{2} - 2xy - x + y = 12...eq2$
subtract eq2 from eq1

${x}^{2} + {y}^{2} + 2xy - x - y - {x}^{2} - {y}^{2} + 2xy + x - y = 56 - 12 \\ \\ 4xy - 2y = 44 \\ \\ 2xy - y = 22 \\ \\ y(2x - 1) = 22 \\ \\ let \: y = 2 \\ \\ 2x - 1 = 11 \\ \\ 2x = 12 \\ \\ x = 6$

So, x= 6 and y = 2

Justification:

$\rm ^{(8)}P_{2} = \frac{8!}{6!} \\ \\ = \frac{8 \times 7 \times 6!}{6!} \\ \\ = 56 \\ \\ \rm ^{(4)}P_{2} = \frac{4!}{2!} \\ \\ = \frac{4 \times 3 \times 2!}{2!} \\ \\ = 12$

Hope it helps you.