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$[x^{50}](x^6 + x^7 + x^8 + .........)^6$

$take \ out \ x^6$

$[x^{50}]x^{36}(1 + x + x^2 + .........)^6$

we have got $x^{36}$, remaining co-efficient is $50 - 36 = 14$

$[x^{14}](1 + x + x^2 + .........)^6$

$[x^{14}]\dfrac{1}{(1-x)^6}$                          $\because \dfrac{1}{1-x} = \sum_{k=0}^{\infty} x^k = 1 + x + x^2 + ...........$

$[x^{14}](1-x)^{-6}$

Now apply $extended \  binomial \  theorem$

=$ \binom{14 + 6 - 1}{14}$                                                    $\because (1-x)^{-n} = \sum_{k=0}^{\infty}  \binom{n+k-1}{k} x^k$

=$\binom{19}{14}$ or $\binom{19}{5}$
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$[x^{50}](x^{6} + x^{7} + x^{8} + ...)^{6}$

$[x^{50}](x^{6}(1 + x^{2} + x^{3} + ... ))^{6}$

$[x^{50}](x^{36})(1 + x^{2} + x^{3} + ... )^{6}$

$[x^{14}](1 + x^{2} + x^{3} + ... )^{6}$

$[x^{14}](\frac{1}{1-x})^{6}$

$[x^{14}](1-x)^{-6}$

$[x^{14}]\binom{-6}{14}(-1)^{14}$

$[x^{14}]\binom{14 + 6 - 1}{14}(-1)^{14}(-1)^{14}$

$[x^{14}]\binom{19}{14}$

$[x^{14}]\binom{19}{5}$

$\binom{19}{5}$ is the answer

Refer https://youtu.be/ZyUb5UxBA9Q

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