Number System Question 2 - CAT 2020

Step-by-Step Explanation

We need to form 4-digit numbers with distinct digits where 7 appears somewhere to the left of 3. Let's break this down into two clean cases based on where the digit 7 is placed:

• Case 1: When 7 occupies the thousands place

If 7 is at the front, 3 can occupy any of the 3 remaining positions (hundreds, tens, or units). The remaining 2 slots can be filled using any of the 8 remaining digits (0, 1, 2, 4, 5, 6, 8, 9) without worrying about zero being at the front.

Arrangements for Case 1 = 3 × 8 × 7 = 168 ways

• Case 2: When 7 is not in the thousands place

Since 7 must come before 3, the digit 3 cannot be in the thousands place either. The thousands place cannot be 0, 7, or 3, leaving 7 choices (1, 2, 4, 5, 6, 8, 9).

Now, out of the 3 remaining positions, choose 2 positions for 7 and 3. Since 7 must be placed before 3, there is only 1 way to arrange them once the positions are chosen ⇒ 3C2 = 3 ways.

The last remaining empty position can be filled by any of the 7 remaining digits (which now includes 0).

Arrangements for Case 2 = 7 × 3 × 7 = 147 ways

Total 4-digit numbers = 168 + 147 = 315

💡 FundaMakers Tip:

Splitting the problem based on the thousands place keeps the number 0 from causing calculation mistakes. Counting options for the leading slot first before placing your restricted digits makes the permutation tracking straightforward.