GATE CS 2010 Q65 — General Aptitude

To form a 4-digit number greater than $3000$ using the digits $2, 2, 3, 3, 3, 4, 4, 4, 4$, the first digit (thousands place) must be either $3$ or $4$.

Case 1: First digit is $3$
The remaining $3$ positions must be filled using the remaining pool of digits: two $2$s, two $3$s, and four $4$s (i.e., $\{2, 2, 3, 3, 4, 4, 4, 4\}$). We count the distinct arrangements based on the selection of these $3$ digits:

• All 3 digits identical: Only $\{4, 4, 4\}$ is possible.
  Number of permutations = $\frac{3!}{3!} = 1$
• 2 digits identical, 1 different:
  • Using two $2$s: $\{2, 2, 3\}$, $\{2, 2, 4\} \implies 2 \times \frac{3!}{2!} = 6$
  • Using two $3$s: $\{3, 3, 2\}$, $\{3, 3, 4\} \implies 2 \times \frac{3!}{2!} = 6$
  • Using two $4$s: $\{4, 4, 2\}$, $\{4, 4, 3\} \implies 2 \times \frac{3!}{2!} = 6$
• All 3 digits distinct: $\{2, 3, 4\} \implies 3! = 6$

Total for Case 1 = $1 + 6 + 6 + 6 + 6 = 25$ distinct numbers.

Case 2: First digit is $4$
The remaining $3$ positions must be filled using: two $2$s, three $3$s, and three $4$s (i.e., $\{2, 2, 3, 3, 3, 4, 4, 4\}$).

• All 3 digits identical: $\{3, 3, 3\}$ and $\{4, 4, 4\}$ are possible.
  Number of permutations = $2 \times \frac{3!}{3!} = 2$
• 2 digits identical, 1 different:
  • Using two $2$s: $\{2, 2, 3\}$, $\{2, 2, 4\} \implies 2 \times \frac{3!}{2!} = 6$
  • Using two $3$s: $\{3, 3, 2\}$, $\{3, 3, 4\} \implies 2 \times \frac{3!}{2!} = 6$
  • Using two $4$s: $\{4, 4, 2\}$, $\{4, 4, 3\} \implies 2 \times \frac{3!}{2!} = 6$
• All 3 digits distinct: $\{2, 3, 4\} \implies 3! = 6$

Total for Case 2 = $2 + 6 + 6 + 6 + 6 = 26$ distinct numbers.

Total distinct numbers = $25 + 26 = 51$.