JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 16)
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 :
$${4 \over {9}}$$
$${9 \over {56}}$$
$${11 \over {27}}$$
$${3 \over {7}}$$
Explanation
Total cases :
$$\underline 6 $$ . $$\underline 6 $$ . $$\underline 5 $$ . $$\underline 4 $$ . $$\underline 3 $$ . $$\underline 2 $$
n(s) = 6 . 6!
Favourable cases :
Number divisible by 3 $$ \equiv $$ Sum of digits must be divisible by 3
Case - I
1, 2, 3, 4, 5, 6
Number of ways = 6!
Case - II
0, 1, 2, 4, 5, 6
Number of ways = 5 . 5!
Case - III
0, 1, 2, 3, 4, 5
Number of ways = 5 . 5!
n(favourable) = 6! + 2 . 5 . 5!
$$P = {{6! + 2.\,5.\,5!} \over {6\,.\,6!}} = {4 \over 9}$$
$$\underline 6 $$ . $$\underline 6 $$ . $$\underline 5 $$ . $$\underline 4 $$ . $$\underline 3 $$ . $$\underline 2 $$
n(s) = 6 . 6!
Favourable cases :
Number divisible by 3 $$ \equiv $$ Sum of digits must be divisible by 3
Case - I
1, 2, 3, 4, 5, 6
Number of ways = 6!
Case - II
0, 1, 2, 4, 5, 6
Number of ways = 5 . 5!
Case - III
0, 1, 2, 3, 4, 5
Number of ways = 5 . 5!
n(favourable) = 6! + 2 . 5 . 5!
$$P = {{6! + 2.\,5.\,5!} \over {6\,.\,6!}} = {4 \over 9}$$
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