Question

Let $S =30 \cdot{ }^{15} C _0+27 \cdot{ }^{15} C _1+24 \cdot{ }^{15} C _2+\ldots \ldots . .12 \cdot{ }^{15} C _{14}-15 \cdot{ }^{15} C _{15}$ has the value $\alpha \cdot 2^\beta$, where $\alpha, \beta \in N$ then the least value of $|\alpha-\beta|$ is equal to

• A. 1
• B. 2
• C. 4
• D. none

Answer 1

Answer: (A)

$ S =30 \cdot C _0 \ldots \ldots-15 \cdot C _{15}$
$S ={ }^{-15} C _0 \ldots \ldots+30 \cdot C _{15}$
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$2 S =15\left( C _0+\ldots \ldots+ C _{15}\right)=15 \cdot 2^{15} $
$\therefore S =15 \cdot 2^{14}$
$\therefore \alpha=15 ; \beta=14$