Question

# The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is -

^{(K + 6)}C_{6} – ^{(K – 4)}C_{6}
^{K}C_{6} – 6 ^{K}C_{4}
^{K}C_{6} – 6 ^{K – 4}C_{6}
^{K + 6}C_{6} – 6 ^{K – 4}C_{6}

^{(K + 6)}C_{6}–^{(K – 4)}C_{6}^{K}C_{6}– 6^{K}C_{4}^{K}C_{6}– 6^{K – 4}C_{6}^{K + 6}C_{6}– 6^{K – 4}C_{6}## The correct answer is: ^{K + 6}C_{6} – 6 ^{K – 4}C_{6}

### The required no. of ways = no. of solution of the equation (x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = K)

Where 0 x_{i} 9, i = 1, 2, …6, where 0 < K < 18

= Coefficient of x^{K} in (1 + x + x^{2} +….. + x^{9})^{6}

= Coefficient of x^{K} in

= Coefficient of x^{k} in (1 – 6x^{10} + 15 x^{20} – ….)

(1 + 6 C_{1}x + 7 C_{2} x^{2} + …. +(7 – K – 10 – 1) C_{K–10} x^{K–10} + ….+(7 + K – 1) C_{K }x^{K} + …)

= ^{k + 6}C_{K} – 6. ^{K–4}C_{K–10}

= ^{k + 6}C_{6} – 6. ^{K–4}C_{6 }.

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^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

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The coefficient of in is

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