The number of 4-digit numbers which can be formed using the digits 0,2,5,7,8 and that are divisible by 2 is
A
300, when repetition is allowed
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
54, when repetition is not allowed
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
60, when repetition is not allowed
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
540, when repetition is allowed
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct options are A300, when repetition is allowed C60, when repetition is not allowed Case 1: When repetition is not allowed. For number to be divisible by 2 the last digit should be 0,2,8. If last digit is 0 then possible numbers =4×3×2=24 If last digit is 2 or 8 ,then possible numbers =3×3×2×2=36 Hence possible numbers will be =36+24=60
Case 2: When repetition is allowed. First place can be filled by either 2or 5 or 7 or 8 in 4 ways Similarly second and thirdplace can be filled by any one of the given digits 5 ways Last place can be filled by either 0 or 2 or 8 in 3 ways. ∴The number of 4 digit numbers divisible by 2 that can be formed using the digits 0,2,5,7,8 when repetition is allowed is =4×5×5×3=300