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Question

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
a17=a16+a15
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B
c17c16+c15
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C
b17b16+c16
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D
a17=c17+b16
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Solution

The correct option is A $$a_{17}=a_{16}+a_{15}$$
In such a number either last digit is $$'0'$$ or $$'1'$$
When you consider $$'a_1'$$, only one number is possible i.e. $$1$$
When you consider $$'a_2'$$, 2 such numbers are possible i.e. $$10$$, $$11$$
When you consider $$'a_3'$$, 3 such numbers are possible i.e. $$101$$, $$111$$, $$110$$
When you consider $$'a_4'$$, 5 such numbers are possible i.e. $$1010$$, $$1011$$, $$1110$$, $$1101$$, $$1111$$
By observing this we get a relationship which is $$a_n$$ $$=$$ $$a_{n-1}$$ $$+$$ $$a_{n-2}$$
So, $$a_{17}$$ $$=$$ $$a_{16}$$ $$+$$ $$a_{15}$$

(Alternate Method)
Using Recursion formula
$$a_n=a_{n-1}+a_{n-2}$$
Similarly, $$b_n=b_{n-1}+b_{n-2}$$ and $$c_n=c_{n-1}+c_{n-2}$$    $$\forall\ n\geq 3$$
and $$a_n=b_n+c_n$$  $$\forall\ n\geq 1$$
So, $$a_1=1, a_2=2, a_3=3, a_4=5, a_5=8 $$ .......
$$b_1=1, b_2=1, b_3=2, b_4=3, b_5=5, b_6=8 $$ .......
$$c_1=0, c_2=1, c_3=1, c_4=2, c_5=3, c_6=5 $$ .......
Using this we get $$b_{n-1}=c_n$$
$$\therefore a_{17}=a_{16}+a_{15}$$

108941_28959_ans_b0e3e3444f8b4fad9baba48c32b5bdd4.png

Mathematics

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