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If $\sum\limits_{ k =1}^{31}\left({ }^{31} C _{ k }\right)\left({ }^{31} C _{ k -1}\right)-\sum\limits_{ k =1}^{30}\left({ }^{30} C _{ k }\right)\left({ }^{30} C _{ k -1}\right)=\frac{\alpha(60 !)}{(30 !)(31 !)}$
Where $\alpha \in R$, then the value of $16 \alpha$ is equal to
$1411$
$1320$
$1615$
$1855$
Solution
$\sum\limits_{ R =1}^{31}{ }^{31} C _{ R } \cdot{ }^{31} C _{ R -1}$
$={ }^{31} C _{1} \cdot{ }^{31} C _{0}+{ }^{31} C _{2} \cdot{ }^{31} C _{1}+\ldots .+{ }^{31} C _{31} \cdot{ }^{31} C _{30}$
$={ }^{31} C _{0} \cdot{ }^{31} C _{30}+{ }^{31} C _{1} \cdot{ }^{31} C _{29}+\ldots .+{ }^{31} C _{30} \cdot{ }^{31} C _{0}$
$={ }^{62} C _{30} \cdot$
Similarly
$\sum\limits_{ R =1}^{30}\left({ }^{30} C _{ R } \cdot{ }^{30} C _{ R -1}\right)={ }^{60} C _{29}$
${ }^{62} C _{30}-{ }^{60} C _{29}=\frac{62 !}{30 ! 32 !}-\frac{60 !}{29 ! 31 !}$
$=\frac{60 !}{29 ! 31 !}\left\{\frac{62 \cdot 61}{30 \cdot 32}-1\right\}$
$=\frac{60 !}{30 ! 31 !}\left(\frac{2822}{32}\right)$
$\therefore 16 \alpha=16 \times \frac{2822}{32}=1411$
