Ans C
Number of four digit numbers from $1001$ to $9999 = 8999.$ Number of times digit group $“37”$ appears is given by:
- $\_ \;\;\_\;\;3 \;\;7 \to \text{(1 to 9) (0 to 9)} \to 9 \cdot 10 = 90$
- $\_\;\;3 \;\;7\; \_ \to \text{(1 to 9) (0 to 9)} \to 9 \cdot 10 = 90$
- $3 \;\;7\;\_\;\; \_ \to \text{(0 to 9) (0 to 9)} \to 10\cdot 10 = 100$
Total count $=90+90+100 = 280.$
Correct option: C
If the question was for finding the number of integers in which $“37”$ appeared then answer would be $279$ because in the above counting we double counted for $3737$, first in $\_ \;\;\_\;\; 3\; \;7$ and second in $3\; \; 7\; \_ \;\;\_ .$