GATE 2020 EE GA Q.8: solution and answer

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

1. A four-digit number $d_1{d}_2{d}_3{d}_4$ has three adjacent pairs of positions where "37" can sit: $(d_1{d}_2)$, $(d_2{d}_3)$ or $(d_3{d}_4)$. Count the occurrences in each position separately, then add them.

2. Position $d_1{d}_2=37$: the numbers are $37\_\_$. Both $d_3$ and $d_4$ are free, giving $10\times 10=100$ numbers (3700 to 3799, all inside the range).

3. Position $d_2{d}_3=37$: the numbers are $d_{1}37d_4$. The leading digit $d_1$ cannot be 0, so it has 9 choices, and $d_4$ has 10: $9\times 10=90$ (1370 to 9379).

4. Position $d_3{d}_4=37$: the numbers are $d_1{d}_{2}37$ with $d_1\ne 0$: $9\times 10=90$ (1037 to 9937). The smallest, 1037, is above 1001, so the lower limit of the range removes nothing.

5. Each occurrence is counted once, in its own position, so the total number of occurrences is $$100+90+90=280$$

6. Check for overlaps: the first and second positions share $d_2$, which would have to be both 7 and 3, so they cannot both hold "37". The same is true of the second and third positions. Only the first and third positions can both hold "37", and that happens in exactly one number, 3737. That number contains the group twice, and both occurrences count.

7. So "37" appears 280 times, spread over $280-1=279$ different numbers. The question asks how many times the group appears, so the answer is 280, option (C).