JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, digits 3, 3, 4, 4, 4, 5, 5 Total 7 digit number from given digits will be
‌

7!

3!2!2!

‌‌=‌

7×6×5×4

2×2

‌‌=210
Now, if the 7 digit number is divisible by 2 , then last digit must be 4 . Now, 7 digit number divisible by 2 will be
=‌

6!

2!2!2!

=‌

6×5×4×3

2×2

=90
Required probability =‌

90

210

=‌

9

21

=‌

3

7