Permutations And Combinations Question 380
Question: If $ ^{n}{C _{r-1}}{{+}^{n+1}}{C _{r-1}}{{+}^{n+2}}{C _{r-1}}+…{{+}^{2n}}{C _{r-1}} $ $ {{=}^{2n+1}}{C _{r^{2}-132}}{{-}^{n}}C _{r}, $ Then the value of r and the minimum value of n are
Options:
A) 10
B) 11
C) 12
D) 13
Correct Answer: C $ \Rightarrow {{}^{n+1}}C _{r}{{+}^{n+1}}{C _{r-1}}+…{{+}^{2n}}{C _{r-1}}{{=}^{2n+1}}{C _{r^{2}-132}} $ $ {{\Rightarrow }^{2n}}C _{r}{{+}^{2n}}{C _{r-1}}{{=}^{2n+1}}{C _{r^{2}-132}} $ $ {{\Rightarrow }^{2n+1}}C _{r}{{=}^{2n+1}}{C _{r^{2}-132}} $ $ \Rightarrow r^{2}-r-132=0 $ $ \Rightarrow (r-12)(r+11)=0\Rightarrow r=12\Rightarrow n\ge 12 $ So, minimum value of n = 12.
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BETA
