Find m and n, if ^{(m+n)}P_{2} = 56 and ^{(m-n)}P_{2} = 12

#### Solution

^{(m+n)}P_{2} = 56

∴ `(("m"+ "n")!)/(("m" + "n" - 2)!)` = 56

∴ `(("m" + "n")("m" + "n" - 1)("m" + "n" - 2)!)/(("m" + "n" - 2)!)` = 56

∴ (m + n)(m + n – 1) = 56

∴ (m + n)^{2} – (m + n) – 56 = 0

Let a = m + n

∴ a^{2} – a – 56 = 0

∴ a^{2} – 8a + 7a – 56 = 0

∴ a(a – 8) + 7(a – 8) = 0

∴ (a – 8)(a + 7) = 0

∴ a – 8 = 0 or a + 7 = 0

∴ a = 8 or a = – 7

∴ m + n = 8 or m + n = – 7

But m + n > 0

∴ m + n = 8 ...(1)

^{(m-n)}P_{2} = 12

∴ `(("m" - "n")!)/(("m" - "n" - 2)!)` = 12

∴ `(("m" - "n")("m" - "n" - 1)("m" - "n" - 2)!)/(("m" - "n" - 2)!)` = 12

∴ (m – n)(m – n – 1) = 12

∴ (m – n)^{2} – (m – n) = 12

Let b = m – n

∴ b^{2} – b – 12 = 0

∴ b^{2} – 4b + 3b – 12 = 0

∴ b(b – 4) + 3(b – 4) = 0

∴ (b – 4)(b + 3) = 0

∴ b – 4 = 0 or b + 3 = 0

∴ b = 4 or b = – 3

∴ m – n = 4 or m – n = – 3

But m – n > 0

∴ m – n = 4 ...(2)

Adding equations (1) and (2), we get,

m + n = 8

m – n = 4

∴ 2m = 12

∴ m = 6

Putting m = 6 in equation (1), we get,

6 + n = 8

∴ n = 8 – 6

∴ n = 2

Hence, m = 6, n = 2