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प्रश्न
Find m and n, if (m+n)P2 = 56 and (m-n)P2 = 12
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उत्तर
(m+n)P2 = 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
∴ a2 – a – 56 = 0
∴ a2 – 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)P2 = 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
∴ b2 – b – 12 = 0
∴ b2 – 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
