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Question

The numbers of integers between 1 and 106 have the sum of their digit equal to K(where 0 < K < 18) is -

  1. (K + 6)C6(K – 4)C6    
  2. KC6 – 6 KC4    
  3. KC6 – 6 K – 4C6    
  4. K + 6C6 – 6 K – 4C6    

The correct answer is: K + 6C6 – 6 K – 4C6


    The required no. of ways = no. of solution of the equation (x1 + x2 + x3 + x4 + x5 + x6 = K)
    Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
    = Coefficient of xK in (1 + x + x2 +….. + x9)6
    = Coefficient of xK in open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent
    = Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
    (1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
    = k + 6CK – 6. K–4CK–10
    = k + 6C6 – 6. K–4C6 .

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