Verified Solution • 2025–26 Session moderate

QJEEAdv12Maa100: JEE Advanced 2012 Mathematics - Class 11 JEEAdv12Maa100 Solution: Let denote the number of all n-digit positive integers formed…

Let an{{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 bn{{b_n}} = the number of such n-digit integers ending with digit 1 and cn{{c_n}} =the number of such n-digit integers ending with digit 0.

Which of the following is correct?

Multiple Choice Options

  • A a17=a16+a15{a_{17}} = {a_{16}} + {a_{15}}
  • B c17≠c16+c15{c_{17}} \ne {c_{16}} + {c_{15}}
  • C b17≠b16+c16{b_{17}} \ne {b_{16}} + {c_{16}}
  • D a17=c17+b16{a_{17}} = {c_{17}} + {b_{16}}

Step-by-Step Verified Solution

For ana_n

Case I : If the unit digit is 1, and rest (n−1)(n-1) places are filled as an−1a_{n-1}

1‾ ‾ ‾ ...‾ ...‾ ‾⏟(n−1) places1‾\mathrm{\underbrace {1\underline {} \,\underline {} \,\underline {} \,\underline {...} \,\underline {...} \,\underline {} }_{(n - 1)\,places}\underline 1}

Case II : If the unit digit is 0, then tenth place must be 1 and rest (n−2)(n-2) places are filled as an−2a_{n-2}

1‾ ‾ ‾ ‾ .... ‾⏟(n−2) place1‾0‾\mathrm{\underbrace {1\underline {} \,\underline {} \,\underline {} \,\underline {} \,....\,\underline {} }_{(n - 2)\,place}\underline 1 \underline 0}

amp; Hence, an=an−1+an−2⇒amp;a17=a16+a15\begin{aligned}& \text { Hence, } a_n =a_{n-1}+a_{n-2} \\\Rightarrow \quad & a_{17} =a_{16}+a_{15}\end{aligned}

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