Permutation and Combination: Which of the following is correct? | JEE Advanced 2012 Paper 2

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

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 $0$.

JEE Advanced 2012 Paper 2Question 1

Which of the following is correct?

Permutation and Combination

$a_{17} = a_{16} + a_{15}$✓ Correct answer$c_{17} \neq c_{16} + c_{15}$$b_{17} \neq b_{16} + c_{16}$$a_{17} = c_{17} + b_{16}$

Solution

$a_n=a_{n-1}+a_{n-2}$ (Fibonacci), so $a_{17}=a_{16}+a_{15}$ is true. Also $c_n=b_{n-1}$ and $b_n=a_{n-1}$ give $c_{17}=c_{16}+c_{15}$ and $b_{17}=b_{16}+c_{16}$, so (B),(C) are false; $a_{17}=b_{17}+c_{17}\ne c_{17}+b_{16}$.

Permutation and Combination - JEE Mathematics question (JEE Advanced 2012 Paper 2)

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