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Q.
If $\frac{\displaystyle\sum_{r=0}^9(-1)^{ r } \cdot 3^{9- r } \cdot{ }^{20} C _{ r } \cdot{ }^{20- r } C _{9- r }}{2^9}={ }^{20} C _{ a }$ then $( a -7)$ can be
Binomial Theorem
Solution:
$\frac{\displaystyle\sum_{ r =0}^9(-1)^{ r } \cdot 3^{9- r } \cdot \frac{20 !}{ r ! \cdot(20- r ) !} \cdot \frac{(20- r ) !}{(9- r ) ! \cdot 11 !}}{2^9}$
$=\frac{\displaystyle\sum_{ r =0}^9(-1)^{ r } \cdot \frac{3^{9- r } \cdot 9 !}{ r !(9- r ) !} \cdot \frac{20 !}{11 ! \cdot 9 !}}{2^9}=\frac{(3-1)^9 \cdot \frac{20 !}{11 ! \cdot 9 !}}{2^9} $
$={ }^{20} C _9 \Rightarrow a =9,11$
$\Rightarrow a -7=2,4$