JEE Advanced · Mathematics · 4. P&C
Let \(a_{n}\) denote the number of all \(n\)-digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . Let \(b_{n}=\) the number of such \(n\)-digit integers ending with digit 1 and \(c_{n}=\) the number of such \(n\)-digit integers ending with digit 0 .
Question:
Which of the following is correct?
- A \(a_{17}=a_{16}+a_{15}\)
- B \(c_{17} \neq c_{16}+c_{15}\)
- C \(b_{17} \neq b_{16}+c_{16}\)
- D \(a_{17}=c_{17}+b_{16}\)
Answer & Solution
Correct Answer
(A) \(a_{17}=a_{16}+a_{15}\)
Step-by-step Solution
Detailed explanation
By recurring formula, \(a_{17}=a_{16}+a_{15}\) is correct
Also \(c_{17} \neq c_{16}+c_{15}\)
\(\Rightarrow a_{15} \neq a_{14}+a_{13}\left(\because c_{n}=a_{n-2}\right)\)
\(\therefore\) Incorrect
Similarly, other parts are also incorrect.
Also \(c_{17} \neq c_{16}+c_{15}\)
\(\Rightarrow a_{15} \neq a_{14}+a_{13}\left(\because c_{n}=a_{n-2}\right)\)
\(\therefore\) Incorrect
Similarly, other parts are also incorrect.
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