Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to :

Solution

Maths sir · 4 steps

The seven digits add to 21, so just ask which digit you leave out, and keep 0 off the front.

1. Leave one digit out. A 6-digit number with no repetition uses all but one of the seven digits. Since 0+1+2+3+4+5+6 = 21, the digit sum is 21 - d, where d is the digit left out.
2. Divisibility test. 21 - d is divisible by 3 only when d is divisible by 3. So the left-out digit must be 0, 3 or 6.

   Digit left out	Digits used	Valid 6-digit numbers	Divisible by 3?
   0	1 to 6	6! = 720	yes
   3	0, 1, 2, 4, 5, 6	6! - 5! = 600	yes
   6	0, 1, 2, 3, 4, 5	6! - 5! = 600	yes
   1, 2, 4 or 5	includes 0	600 each	no

3. Count both. Favourable: 720 + 600 + 600 = 1920. Total: the first digit has 6 choices (not 0), then 6 × 5 × 4 × 3 × 2 for the rest, giving 6 × 720 = 4320.

   P(A) = 1920/4320 = 4/9

4. So. The probability of event A is 4/9, which is option (a).

The idea to keep. For divisibility by 3 with a fixed set of digits, the order does not matter, only the digit sum. Find which digits to drop, then count arrangements with 0 kept out of the first place.

Answer from the NTA final answer key. Solution by PrepWiser.