Question

How many five-digit numbers formed using
the digit 0, 1, 2, 3, 4, 5 are divisible by 5 if
digits are not repeated?
​

Answer 1

Answer:

The numbers are divisible by 5 is they end with either 0 or 5

★Case 1

Let the number ends with 0 i.e. unit's digit is 0

Now there are 5 choices for ten's place

As the repetition is not allowed, 4 choices for hundred' place.

3 choices for thousand's place and 2 choices for the first digit.

So, the 5 digits numbers that can be formed which ends with 0 is 5×4×3×2=120

★Case 2:

Let the number ends with 5 i.e. unit's digit is 5

As the repetition is not allowed, 4 choices for hundred' place.

3 choices for thousand's place and 2 choices for the first digit.

So, the 5 digits numbers that can be formed which ends with 5 is 5×4×3×2=120

But the first digit can't be 0. So, we need to subtract that numbers which contains 0 as first digit.

If 0 is the first digit, then number of places left to be filled is 4 and that can be done in 4! ways.

So, the 5 digits numbers that can be formed which ends with 5 is 120−4!=96

Thus, the total number of five digit numbers divisible by 5 is 120+96=216

Step-by-step explanation:

in simple words-

{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set. {0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set. {1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set. Total = 5*5*4*3*2 = 600 but the numbers which are divisible by 3 come from third and sixth sets: 96 + 120 = 216.

Thus, the total number of five digit numbers divisible by 5 is 120+96=216.

have a nice day...

Answer 2

Answer:

{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set. {0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set. {1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set. Total = 5*5*4*3*2 = 600 but the numbers which are divisible by 3 come from third and sixth sets: 96 + 120 = 216.

profile is very cute

me follow