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Find m and n, if (m+n)P2 = 56 and (m-n)P2 = 12 - Mathematics and Statistics

Sum

Find m and n, if (m+n)P2 = 56 and (m-n)P2 = 12

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Solution

(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

Concept: Concept of Permutations - Permutations When All Objects Are Distinct
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 3 Permutations and Combination
Exercise 3.3 | Q 2 | Page 54
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