IIT-JEE 2012 · Paper 2 · Mathematics Q50

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Paragraph for Questions 49 and 50

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.

Which of the following is correct?

a_{17}=a_{16}+a_{15}

c_{17}\ne c_{16}+c_{15}

b_{17}\ne b_{16}+c_{16}

a_{17}=c_{17}+b_{16}

Answer and worked solution

Official answer: A

From $b_n=a_{n-1}$ and $c_n=b_{n-1}=a_{n-2}$ :
$a_n=b_n+c_n=a_{n-1}+a_{n-2}.$

(A) $a_{17}=a_{16}+a_{15}$ . True.
(B) $c_n=a_{n-2}$ also satisfies the Fibonacci recurrence, so $c_{17}=c_{16}+c_{15}$ . False.
(C) $b_{17}=a_{16}=b_{16}+c_{16}$ . False.
(D) $c_{17}+b_{16}=a_{15}+a_{15}=2a_{15}\ne a_{17}$ . False.

(A)

Topics:Fundamental principle of counting

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IIT-JEE 2012 · Paper 2, Mathematics question 50