JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Permutations and Combinations

A number is divisible by 6 if it is divisible by both 2 and 3.

For divisibility by 3, the sum of digits must be divisible by 3. Sum of all six digits = $1+2+3+5+6+7=24$. We need to choose 5 digits whose sum is divisible by 3:

• Exclude 3: $1,2,5,6,7$ (sum = 21, divisible by 3) ✓
• Exclude 6: $1,2,3,5,7$ (sum = 18, divisible by 3) ✓

For divisibility by 2, the last digit must be even.

Case 1: Using $1,2,5,6,7$. Even digits available: 2, 6.
Number of arrangements = $4!\times 2=24\times 2=48$.

Case 2: Using $1,2,3,5,7$. Even digit available: 2.
Number of arrangements = $4!\times 1=24$.

Total = $48+24=72$.