Question

Heya... brainliacs ....

How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3, 4, 5 if 

(i) repetition of digits are not allowed? 

(ii) repetition of digits are allowed?

Answer 1

No Formula used -







By divisibilty rule of 5, we came to know that a number will be divisible by 5 only when the number will have 0 or 5 at its ones place.



Now,

( i ) numbers starting from 1

105,120,125,130,135,140,145,150

Numbers formed : 7



So, total numbers will be formed by putting 1,2,3,4 = 4( 8 ) = 32


( ii ) : numbers from 5 [ which is last ]

510,520,530,540

Numbers = 4





Hence, total numbers = 32 + 4 = 36







( ii )

Repetition is allowed,


Numbers forming by taking 1 as hundreds place,


( i ) 100,105,110,115,120,125,130,135,140,145,150,155


Numbers formed = 12





There last term will not contain less or more numbers as repetition is allowed, so


Total numbers = 5( 12 ) = 60
$.$

Answer 2

Given, numbers: - 0,1,2,3,4,5
To form number of 3 digits We have 3 places!
(I) [text] \textbf{Refer to attachment} [\text]

(ii) if repeation is allowed

*Number of way to fill first place = 2
(We can only fill 0 or 5 for divisibility by 5 )

*Number of ways to fill second place = 6

*Number of ways too fill third place = 5
(we can not fill 0 at third)

Hence,
Total number of way = 2×5×6 = 60