**Counting procedure of a Johnson's Counter**

- Starting from all 0's , each shift operation inserts 1's from the left until the register is filled with all 1's.
- When the register has all 1s,each shift operation inserts 0's from the left until the register is filled with all 0's.
- Go to step 1.

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0000 ** $\rightarrow$** ${\color{Red} 1}$000** $\rightarrow$** ${\color{Red} 1}{\color{Red} 1}$00 **$\rightarrow$** ${\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$0** $\rightarrow$** ${\color{Red} 1}{\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$ ** $\rightarrow$ ** ${\color{Blue} 0}{\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$** $\rightarrow$** ${\color{Blue} 0}{\color{Blue} 0}{\color{Red} 1}{\color{Red} 1}$ **$\rightarrow$** ${\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}{\color{Red} 1}$** $\rightarrow$** ${\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}$

**OR**

0 ** $\rightarrow$** 8 ** $\rightarrow$** 12 ** $\rightarrow$** 14 ** $\rightarrow$** 15 ** $\rightarrow$** 7 ** $\rightarrow$** 3 ** $\rightarrow$** 1 ** $\rightarrow$** 0

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$\therefore$ Option D is correct answer.