A) 0
B) \[\frac{{{2}^{20}}-1}{2}\]
C) \[{{2}^{19}}\]
D) \[\frac{{{3}^{20}}-1}{2}\]
Correct Answer: A
Solution :
\[\frac{20!}{p!q!r!}{{(2x)}^{p}}{{(-y)}^{q}}{{(z)}^{r}}=\frac{20!}{p!q!r!}{{2}^{p}}{{(-1)}^{q}}{{x}^{p}}{{y}^{q}}{{z}^{r}}\]\[p+q+r=20,q=0\] p + r = 20 (p is even and r is odd). even + odd = even (never possible) Coefficient of such power never occur \[\therefore \]coefficient is zero
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