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The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, z less or equal than –1

  1. 31    
  2. 30    
  3. 29    
  4. None of these    

The correct answer is: 30


    Let t = z + 1
    Equation reduces to x + y + t = 25
    less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, t greater or equal than 0
    Required number of ways
    = Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
    = Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
    = Coefficient of x12 in fraction numerator left parenthesis 1 minus x to the power of 6 end exponent right parenthesis over denominator left parenthesis 1 minus x right parenthesis end fraction (1 – x)–1
    = Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
    = Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
    = 12+3–1C3–17+3–1C3–16+3–1C3–1 + 1+3–1C3–1
    = 14C29C28C2 + 3C2
    = 91 – 36 – 28 + 3 = 30.

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